q q 1 mv 2 2 2 . . . . . . . 1 mv2 0 2 . . v . 0 The particlewill just reach centre of ring. Example-15 : Find the electric field at centre of semicircular ring shown in figure. ����� � � + +++++ Y R X �q q Sol. . 2 2 0 0 0 E . E . E . 2 E But 0 0 E sin 2 R 4 . . . . . But 0 K 4 l . . . , 2 2q R . . . ����� � � + ++ +++ 45º 45º 45º �q +q E0 E0 O X E Y 45º . 2 E 4Kq .R . directed negative x-axis. Example-16 : Asimple pendulumof length l and bobmassmis hanging in front ofa large nonconducting sheet having surface charge density.. Ifsuddenlya charge +q is given to the bob&it is released fromthe position shown in figure. Find the maximum + + +

+ + ++ + + l s angle throughwhich the string is deflected fromvertical. Sol. The electric field intensitydue to nonconducting sheet is 0 E 2. . . ...(1) According towork energy theorem Wnet = .k = 0 Since initial and finalvelocity ofbob is zero Therefore, WE + Wg = 0 ...(2) WE = qE . sin . Wg = �mg ( . � . cos.) Putting the values in eqn (2) + + ++ + ++ + + cosq s mgv=0 qE qsinq qE . sin. �mg. (1�cos.) = 0 qE . sin. =mg. (1�cos.) qE 1 cos mg sin . . . . . . . 2 sin2 qE 2 mg 2sin cos 2 2 . . . . qE tan mg 2 . . . tan 1 qE 2 mg . . . . . . . . .

2 tan 1 qE mg . . . . . . . . . Putting the value ofE fromeqn (1) 1 0 2 tan q 2 mg . . . . . . . . . . . . 1 02 tan q 2 mg . . . . . . . . . . . Ans.

ELECTROSTATICS www.physicsashok.in 15 Example-17 : A particle ofmassmand negative charge q is thrown in a gravityfree spacewith speed u fromthe pointAon the large non conducting charged sheet with surface charge density ., as shown in figure. Find themaximumdistance fromAon sheet where the particle can strike. + + ++ + ++ + u A Sol. Electric field intensitydue to nonconducting sheet 0 E 2. . . ...(1) where ux = u sin. uy = u cos. x a qE m = + + ++ + ++ + u A ucos usin X Y 2 x x x u T 1 a T 2 = + where x = 0 2 x x 0 u T 1 a T 2 = + . 0 .u sin .T 1 qE T2 2 m . . . .u sin .T 1 qE T2 2 m . . . T 2umsin qE

.

. y = uyT y= (u cos.)T y u cos 2u msin qE . . . . . y u2msin 2 qE . . y 2 0 2u m sin 2 q . . . . fromeqn (1) . 2 0 y 2u msin 2 q . . . . 2 0 max y 2u m q.. . At ymax, sin2. = 1 Example-18 : The figure shows three infinite non-conducting plates of charge perpendicular to the plane of the paper with charge per unit area +., +2. and �., Find the ratio of the net electric field at that pointAto that at point B. + + � + + � + + � + + � + + � + + � + + � A B 2.5m 2.5m 5 m 5 m +s +2s �s Sol. The electric field intensity at point Adue to plate x, y and z EA = Ex + Ey + Ez 0 0 0 2 2 2 2 . . . . . . . .

EA = 0 ...(1) A B +s +2s �s x y z 2s 2Î 0 s 2Î 0 s Î 0 At point B B 0 0 0 E 2 2 2 2 . . . . . . . . . B 0 E 4 2 . . . ...(2) Fromeqn (1) and (2) AB E 0 E =

ELECTROSTATICS www.physicsashok.in 16 Example-18 : A metallic solid sphere is placed in a uniformelectric field. The lines of force follow the path(s) shown in figure as : (A) 1 (B) 2 1234 432 1 (C) 3 (D) 4 Sol. Electric field lines never enter a metallic conductor (E = 0, inside a conductor) and they fall normally on the surface of a metallic conductor (because whole surface is at same potential and lines are perpendicular to equipotential surface). Example-19 : Anon-conducting solid sphere of radius Ris uniformly charged. Themagnitude of electric field due to the sphere at a distance r from its centre : (A) increases as r increases for r < R (B) decreases as r increases for 0 < r < . (C) decreases as r increases for R < r < . (D) is discontinuous at r = R (E) both (a) and (c) are correct Sol. 0 E r 3.. . r < R E .r r < R 2 0 E Q 4.. r . r > R 2 E 1 r . Hence, both (a) and (c) are correct. Example-20 :Aparticle ofmassmand charge �qmoves along a diameter of a uniformlycharged sphere of radius Rand carrying a total charge+Q. Find the frequencyofS.H.M. of the particle if the amplitude does not exceed R. Sol. Electric field intensitydue to nonconducting sphere 0 E x 3. . . ...(1) (where x < R) Where . is volume charge density. The force onnegative charge is opposite direction of electric field. �qE = ma Fromeqn (1) 0

q x ma 3. . . . . 0 a q x 3 m . . . . ...(2) xR �q For S.H.M. a = �.2x ...(3) Fromeqn (2) and (3) 2 0 q 3 m . . . . . 0 q 3 m . . . . 0 2 f q 3m. . . . {. . . 2.f } 0 1 q f 2 3m. . . . . 3 0 1 qQ f 2 4 mR . . .. 3 Q 4 / 3 R . . . . . . . . . . Example-21 : Apositive chargeQis uniformlydistributed throughout the volume of a dielectric sphere of radius R.Apointmass having charge +q andmassmis fired towards the centre of the spherewith velocityv froma point at distance r (r >R) fromthe centre of the sphere. Find theminimumvelocityvso that it canpenetrateR/

ELECTROSTATICS www.physicsashok.in 17 2 distance of the sphere. Neglect any resistance other than electric interaction. Charge on the smallmass remains constant throughout themotion. Sol. Frommachenicalenergyconservation U i + T i = Uf + Tf qV i + ½mv2 = qVf + 0 ...(1) i 0 V Q 4. r . . , 2 2 f 3 0 1 Q(3R r ) V 2 4. R . . . where r R2 = R O vmin 2R r +q 2 2 f 3 0 Q 3R R1 4 V 2 4. R . . . . . . . . . . . 2 2 . f 3 0 Q 12R R V 32. R . . . f 0 V Q(11) 32. R

.

. . f 0 V 11Q 32. R . . Putting the values ofVi andVf ineqn. (1) 2 0 0 qQ 1 mv 11qQ 4. r 2 32. R . . . . . 2 0 0 1 mv 11qQ qQ 2 32. R 4. r . . . . 2 0 0 mv 11qQ qQ 16. R 2. r . . . . . 2 0 mv qQ 11 1 2. 8R r . . . . . .. .. 2 0 v qQ 11 R 2m. R 8 r . . . . . .. .. . v2 2kqQ 1 3 R mR 8 r . . . . . .. .. v2 2kqQ r R 3 mR r 8 . . . . . . . . . . 12kqQ r R 3 2 v mR r 8 . . . .. . . . . .. . . .. Example-22 : The diagramshows a small bead ofmassmcarrying charge q. The beamcan freelymove on the smooth fixed ring placed on a smooth horizontal plane. In the same plane a charge +Qhas alos been fixed as shown. The potential at the point P due C4a a+Q B xg P to +Q isV. The velocitywithwhich the bead should projected fromthe point P so that it can complete a circle should be greater than

(A) 6qV m (B) qV m (C) 3qV m (D) none Sol. According to conservationprincipalofmechanical energy. Ui + Ti = Uf + Tf 20 0 qV 1 mv qQ 0 2 4. a . . . . ...(1) where v0 is velocityof at point �p� the potential at the point p due to +Qis C4a a+Q B xg P v0 0v Q 4. (4a) . . fromeqn (1)

ELECTROSTATICS www.physicsashok.in 18 20 qv 1 mv 4qV 2 . . 201 mv 3qV 2 . 20v 6qV m . . 0 v 6qV m = Ans. Example-23 : Two spherical, nonconducting, and verythin shells of uniformlydistributed positive chargeQand radius d are located a distance 10d fromeach other.A positive point charge q is placed inside one of the shells at a distance d/2 from d/2 10 d d Q Q the center, on the line connecting the centers of the two shells, as shown in the figure. What is the net force on the charge q ? (A) 2 0 qQ 361.. d to the left (B) 2 0 qQ 361.. d to the right (C) 2 0 362qQ 361.. d to the left (D) 2 0 360qQ 361.. d to the right Sol. Electric force on charge q due to sphreAis zero. But electric force due to sphere B on charge q is 2 0 F qQ towards left 4 19 d 2 .. . . . .. .. Q Q B q F 19 d 2

A d/2 2 0 qQ towards left 361.. d . Example-24 : The diagramshows three infinitely long uniformline charges placed on the X, Yand Z axis. The work done in moving a unit positive charge from(1, 1, 1) to (0, 1, 1) is equal to Y X Z l 3l 2l (A) (. ln 2) /2..0 (B) (. ln 2)/..0 (C) (3. ln 2) /2..0 (D) None Sol. Here r = 1+ x2 E1 = The magintude of electric field due to wire along y-axis = 2 03 2 1. x . .. directed paerpendicular to y-axis E2 = The magnitude of electric field due to wire 1+ x2 E along z-axis = 2 1 x l+ directed perpendicular to z-axis. The electric field due to wire along x-axis is directed perpendicular to x-axis. . The net component of electric field along x-axis is x 2 2 0 0 E 3 cos cos 2 1 x 2 1 x . . . . . . . . .. .. where . = angle made by E1 with x-axis and . = angle made by E2 with x-axis But 2 cos cos x 1 x . . . . . . x . 2 . 0E 4 x 2 1 x . . . ..

ELECTROSTATICS www.physicsashok.in 19 . �dV = Ex dx or . . x 0 2 x 1 0 dV 4 x dx 2 1 x .. . . . . . . .. 21 v x 0 2 v 0 x 1 dV 4 x dx 2 1 x .. . . . . . . .. x 0 1 2 2 0 x 1 V V 4 x dx 2 1 x .. . . . . . .. Put 1 + x2 = z or 2x dz dx = or dz = 2 xdx x 0 1 2 0 x 1 V V 2 x dz z 2x .. . . . . . .. x 0 0 x 1 dz z .. . . . ..

x 0 x 1 0 lnz .. . .. .. . .. 01 0 . 2 ln(1.x) .. .. . .. 01 0 . ..ln1.ln2.. . ..2 1 0 v . v . ln 2 . .. Example-25 : A particle ofmass 1 kg & charge 1 µC 3 is projected towards a non conducint fixed spherical shell having the same charge uniformly V from 1 mm distributed on its surface. Find the minimuminitial velocity of projection required if the particle just grazes the shall. (A) 2 m/ s 3 (B) 2 2 m/ s 3 (C) 2 m/ s 3 (D) none of these Sol. Apply conservation principle of angular momentum, m vd = mv0 r . 0 v vd v r 2 . . d r 0.5 mm 2 . . . . ... .. Applying mechanical energy conservation principle. Ui + Ti = Uf + Tf 2 2 20 0 1 q 1 0 mv mv 2 4 r 2 . . . ..

or 2 2 20 0 1 q 1 mv mv 2 4 r 2 . . ..

ELECTROSTATICS www.physicsashok.in 20 or 2 2 2 0 1 mv q 1 m v 2 4 r 2 2 . . . . .. .. .. or 2 2 2 0 1 mv q mv 2 8 4 r . . .. or 2 2 0 1 3 q mv 2 4 4 r . . .. or 2 2 0 1 q mv 2 3 r . .. or 2 2 0 8q v 3 4 mr . . .. 9 12 3 8 9 10 1 10 9 8 3 1 1 10 3 . . . . . . . . . . . v 8 2 2 m/ s 3 3 = = ELECTRIC POTENTIAL ENERGY Two like charges repel each other while the two unlike charges attract each othe

r. So, when the charges are moved awayfromeach other or they are brought near each other, either somework is obtained or somework is done.Thiswork is stored in the systemofcharges in formof electric potential energy. �The electric potential energyof a systemof different point charges is equal to thework done inbringing those charges frominfinityto formthe system.� It is represented byU. The electric potentialenergyof a systemof two point charges q1 and q2 invacuumat a separation r is given by, 1 2 0 U 1 q q J 4 r . .. Electric Potential Energy of a System of Charges : The electric potential energy of a systemof charges is given by i j i j 0 ij 1 q q U J 4 . r . . .. This sumextends over allpairs of charges.We do not let i=j, because that would be an interaction ofa charge with itself andwe include onlytermswith i < j to make sure that we count each pair only once. For example, electric potential energyof four point charges q1, q2, q3 and q4 would be given by, 4 3 4 2 4 1 3 2 3 1 2 1 0 43 42 41 32 31 21 1 q q q q q q q q q q q q U 4 r r r r r r . . . . . . . . . . .. . . Here, all the charges are to be substitutedwith sign. NOTE : Total number of pairs formed by n point charges are n.n - 1. . 2 Example-26 : Four charges q1 = 1µC, q2 = 2µC, q3 = �3µC and q4 = 4µC are kept on the vertices of a square of side 1m. Find the electric potential energy of this systemof charges. q4 q3 q1 q2 1 m 1 m Sol. In this problem,

ELECTROSTATICS www.physicsashok.in 21 r41 = r43 = r32 = r21 = 1m and 2 2 r42 . r31 . (1) . (1) . 2 m Substituting the proper valueswith signin the relation givenabove in the theory,we get U = (9.0 × 109) (10�6) (10�6) (4)( 3) (4)(1) (4)(1) ( 3)(2) ( 3)(1) (2)(1) 1 2 1 1 2 1 . . . . . . . . . . . . . . . .9.0 10 3 . 12 52 . . . .. . . .. .. = �7.62 × 10�2 J NOTE : Here negative sign of U implies that positive work has been done by electrostatic force in assembling these charges at respective distances from infinity. Example-27 : Consider the configurationof a systemof four charges each of value +q. Find thework done byexternal agent in changing the +q +q +q +q a a fig (i) fig (ii) +q +q +q +q a configuration of the systemfromfigure (i) to fig (ii). Sol. Ui =Electricalpotential energyinsquare arrangement 2 2 i 0 0 U 4q 2q 4 a 4 2a . . . . . . Uf = Electric potential energyof circular arrangement of charge q a2 a a q q q 2 2 f 0 0 U 4q 2q 4 2a 4 (2a) . . . . . . ext w . .U . . 2 f i 0 q U U 3 2 4 a

. . . . .

. . Example-28 : Aconemade of insulatingmaterialhas a total chargeQspread uniformly over its sloping surface.Calculate the energyrequired to take a test charge q frominfinity to apexAof cone. The slant length is L. B A AB = L Sol. ext f i w . .U . U . U A q(V V ) . . . wext A A . q(V . 0) . qV ...(1) HereVA = Electric potential at pointA. The electric charge on the considered ring dq Q2 rdx RL . . . . . Electric potentialdue to consideredring is 0 dV dq 4 x . . . . 0 dV Q2 rdx RL4 x . . . . . . sin r R x L . . . Ar dx q x

ELECTROSTATICS www.physicsashok.in 22 . r Rx L . . 0 2 Q R x dx dV L RL4 x . . .. .. . . . . . . L A 2 0 0 V dv Q dx 2 L . . . . . . 0 Q L 2 = p Î . ext A 0 w q V qQ 2 L . . . . fromeqn (1) RELATION BETWEEN ELECTRIC FIELD AND POTENTIAL In case of cartesian co-ordinates x y z E . E i� . E �j. E k� .. Here, x E Vx . . . . . . (partialderivative ofVw.r.t. x) y E Vy . . . . . . (partialderivative ofVw.r.t. y) z E Vz . . . . . . (partial derivative ofVw.r.t. z) . E V �i V �j V k�

x y z . . . . . . . . . . . . . . . . .. This is alsowritten as, E . .grad V . ..V .. .. Example-29 : The electric potential in a region is represented as, V= 2x + 3y� z, Obtain expression for electric field strength. Sol. E V �i V �j V k� x y z . . . . . . . . . . . . . . . . .. Here, . . V 2x 3y z 2 x x . . . . . . . . V .2x 3y z. 3 y y . . . . . . . . V .2x 3y z. 1 z z . . . . . . . . . . E . .2 i� . 3�j. k� .. In polar co-ordinates r E Vr . . . . and E 1 . V r . . . . .. For example, electric potential due to a point charge q at distance r is 0 V 1 . q 4 r . .. . E. = 0 V 0 . . . . . . . .. . . and r 2 0 E 1 . q 4 r . ..

ELECTROSTATICS www.physicsashok.in 23 Thus, electric field E dV dr . . or E = � (slope of v�r graph) NOTE : � If electric field .. E is known, then electric potential can be determined from the relation as given below: . d .. . dV = -E r or . . d b .. . b a a V -V = - E r Here, dr . dx �i . dy �j. dz k� . � In uniform electric field V = Ed Example-30 : FindVab in an electric field E .2 i� 3�j 4 k� . NC . . . .. where . . a r . i� . 2�j. k� m . and . . b r . 2�i . �j. 2k� m . Sol. Here, the given field is uniform(constant). So using, dV . .E . dr .. . or a ab a b b V . V . V . .. E . dr .. .. . . . . . .1, 2, 1. 2, 1, 2 2�i 3�j 4k� . dx �i dy�j dz k� . . . .. . . . . . . . . .1, 2,1. 2,1, 2 2dx 3dy 4dz . . . .. . . . .. . .1, 2, 1. 2,1, 2 2x 3y 4z . . . . . . = � 1 V Example-31 : Auniformelectric field having strengthE.. is existing in x�yplane as shown in figure. Find the p.d. between originO &A(d, d, 0)

(A) Ed(cos. + sin.) (B) �Ed(sin. � cos.) E A(d,d,0) o x y z (C) 2 Ed (D) none of these Sol. Here E . Ecos. �i . Esin. �j .. . OA . d �i . d �j . .... . ..v . E . .. .. ..v . Ed cos. . Ed sin. . Ed(cos. . sin. ) A 0 . . (V .V ) . Ed(cos. . sin.) 0 A . V . V . Ed(cos. . sin. ) Example-32 : Uniformelectric field ofmagnitude 100V/min space is directed along the line y= 3 + x. Find the potential difference between pointA(3, 1)&B (1, 3) (A) 100 V (B) 200 2 V (C) 200 V (D) 0 Sol. . y = 3 + x . tan. = 1

ELECTROSTATICS www.physicsashok.in 24 . . = 45º E .100cos. �i .100sin. �j .. E 100 �i 100 �j 2 2 . . .. .r . AB . i� . 3�j. 3�i . �j . .2�i . 2�j ... .... .V . .E . .r .. ... 100 � 100 � V i j r 2 2 . . . . .. . . . . . . ... V 100 �i 100 �j . 2�i 2�j. 2 2 . . . . .. . . . . . . . . .V . .100 2 .100 2 . A B V - V = 0 Example-33 : A, B, C, D, P andQare points in a uniformelectric field. The potentials a these points are V (A) = 2 volt. V(P) =V(B) =V(D) = 5 volt. V(C) = 8 volt. The electric field at P is 0.2m AB P CQD 0.2m (A) 10 Vm�1 along PQ (B) 15 2 Vm�1 along PA (C) 5 Vm�1 along PC (D) 5Vm�1 along PA Sol. A D x E V V 2 5 30 15 v /m 0.2 0.2 2 . . . . . . . . Also A B y E V V 2 5 30 15 v /m AB 0.2 2 . . . . . . . . E = - 15�i - 15�j .. Py x 15 15 x�A 45º

| E |= (- 15)2 + (- 15)2 = 15 2 v /m .. PA E .15 2 Hence (B) is correct. Example-34 : Variation of electrostatic potential along x-direction iswhosn in the graph. The correct statement about electric field is (A) xcomponent at point Bismaximum (B) x component at pointAis towards positive x-axis x A B C v (C) x component at point C is along negative x-axis (D) x component at point C is along positive x-axis Sol. The negative slope v � x graph give x � component of electric field. In the given graph, slope at C is negative. Hence, x � component of electric field is positive. EQUIPOTENTIAL SURFACE Equipotential surface is an imaginarysurface joining the points of same potential inan electric field. So,we can say that the potential difference between anytwo points on anequipotential surface is zero.The electric lines of force at each point ofan equipotential surface are normal to the surface. Fig. shows the electric lines of force due to a point charge +q. +q This spherical surfacewillbe the equipotential surface and the electrical lines of force emanating fromthe point chargewill be radial and normal to the spherical surface.

ELECTROSTATICS www.physicsashok.in 25 Regarding equipotentialsurface, following points areworth noting : (a) Equipotential surfacemaybe planar, solid etc.But equipotential surface never be a point. (b) Equipotential surface is single valued. So, equipotential surfaces never cross each other. (c) Electric field is always perpendicular to equipotential surface. (d) electric lines of force cross equipotentialsurface perpendicularly. (e) Work done to move a point charge q between two points on equipotential surface is zero. (f) The surface ofa conductor in equilibriumis equipotential surface. (g) equipotential surface due to isolated point charge is spherical. (h) Equipotential surface are planar in uniformelectric field. (i) Equipotential surface due to line charge is cylindrical. (j) Equipotential surface due to an electric dipole is shown inthe figure. �q +q Example-35 : The electric field in a region is givenby : E . .4axy z .�i . .2ax2 z .�j. .ax2y / z .k�, where a is a positive constant.The equation of anequipotential surfacewill be of the form (A) z = constant/[x3y2] (B) z= constant/[xy2] (C) z = constant/[x4y2] (D) none Sol. At equipotential surface potential is constant. Therefore E v r = - d d .. dv . . E . dr .. . . E .d r . constant .. . r . x i� . y�j. z k� . . dr . dx�i . dy�j. dzk� . . E . (dx �i . dy�j. dz k�) .. = Constant

ELECTROSTATICS www.physicsashok.in 26 .(4axy z )i� . (2ax2 z)�j. (ax2y / z)k� . . (dx�i . dy�j. dzk�) . . . = constant 2 4axy z dx 2ax2 z dy ax ydz z . . . . . = constant. 2 4ax y z 2ax2y z 2ax2y z 2 . . = constant 6 ax2 y z = constant 2 z constant 6ax y = .. 4 2 z constant x y . Example-36 : The equation of an equipotential line an electric field is y= 2x, thenthe electric field vector at (1, 2) may be (A) 4�i +3�j (B) 4�i +8�j (C) 8i� + 4�j (D) �8�i + 4�j Sol. Electric field is perpendicular to equipotential line . y = 2 x or dy 2 dx = . m1 = 2 . m1m2 = � 1 . 2 m 12 = - . tan 12 . . . where . is anglewith x-axis. In option (D), tan 4 1 8 2 . . . . . ELECTRIC DIPOLE Electric dipole is a systeminwhich two equal and opposite point charges are placed at a smalldistance. The product of any of the charges and distance between two charge is called electric dipole 2l �q P +q moment p. It is directed fromnegative charge to positive charge (fig.). The line joining the two charges is called axis of dipole. Let charges of an electric dipole are �q and +q and are separated by a small distance 2l. The dipolemoment of such a dipole is given by p = q × 2l = 2ql Electric Potential and Field due to an Electric Dipole Let an electric dipole is placed along y-axis and its centre is at origin, thene

lectric potential at pointA(x, y, z) due to this dipole. +q�q z y l l x2 + z2 + (y+l)2 x2 + z2 + (y�l)2 A(x, y, z) x

ELECTROSTATICS www.physicsashok.in 27 2 2 2 2 2 2 0 1 q q V 4 x (y ) z x (y ) z . . . . . . .. .. . . l . . . l . .. x E Vx . . . . , y E Vy . . . . and z E Vz . . . . Special Cases : (i) On the axis of dipole (axial position) : x = 0, z = 0 (a) 2 2 0 V 1 p 4 (y ) . .. . l 2 2 0 V 1 p 4 (r ) . .. . l If y = r or axis 2 0 V 1 P 4 r . .. If r > > l (b) Ex = 0 = Ez and y 2 2 2 0 E E 1 2py 4 (y ) . . .. . l (alongp . ) 2 2 2 0 1 2pr 4 (r ) .

.. . l If y = r axis 3 0 E 1 2p 4 r . .. (ii) On the perpendicular bisector of dipole (equatorical position) : Say along x-axis (it may be along z-axis also) y = 0, z = 0 (a) bisec tor V 0 . . (b) Ex = 0, Ez = 0 and y 2 2 3/ 2 0 E 1 p 4 (x ) . . .. . l (opposite to p . ) Here, negative sign implies that the electric field is along negative y-direction or antiparalleltop . . Further at a distance r fromthe centre of dipole (x= r), then 2 2 3/ 2 0 E 1 . p 4 (r ) . .. . l or bisec tor 3 0 E 1 p 4 r . . .. if r >> l (iii) In polar co-ordinates (r, .) : (a) 2 0 V 1 p cos 4 r . . .. (b) r E Vr . . . . 3 0 1 . 2pcos 4 r . . .. E 1 . V r . . . . .. 3 0 1 psin

4 r . . .. E Er E P r O A(r, ) . 2 2 r E E E. . . . 2 3 0 E p 1 3cos 4 r . . . .. (c) In Figure, tan tan2 . . .

ELECTROSTATICS www.physicsashok.in 28 Example-37 : 4 charges are placed each at a distance �a� fromorigin. The dipolemoment of configurationis (A) 2qa�j (B) 3qa�j 3q�2q �2q qy x (C) 2aq[i� + �j] (D) none Sol. Here 1 P . 2q . 2 a 1 P = 2 2 qa 2 P . q . 2a . 2 qa 3 P . q . 2a . 2 qa 3q �2q �2q q y x P1 45º 45º 45º P3 P2 2a2a 2a . x 1 2 3 P = P cos 45 - P cos 45 - P cos 45 x P = 2 2qa cos 45 - 2qa cos 45 - 2qa cos 45 x P = 0 y 1 2 3 P = P sin 45 + P sin 45 - P sin 45 y 2 2qa 2qa 2qa P 2qa 2 2 2 = + - = . P . 2 q a�j .. Example-38 : Two point charges +3 µC and �3 µC are placed at a small distance 2 × 10�3 mfromeach other. Find: (a) electric field and potential at a distance 0.6mfromdipole in its equatorialposition. (b) electric field and potential at the same distance fromdipole as in (a) if the dipole is rotated through 90º. Sol. Dipolemoment p = q × 2l = (3 × 10�6) (2 × 10�3) p = 6 × 10�9 C-m (a) Electric field in equatorialposition, 3 0 E 1 . p 4 r . .. . . 9

9 3 E 9 10 6 10 (0.6). . . . . . . . . . E = 250 N/C Electric potential,V= 0 (b) On rotating the dipole through 90º, the same point nowwill be in axial position. So, electric field 3 0 E 1 . 2p 500 N/C 4 r . . .. and electric potential 2 0 V 1 p 4 r . .. . . 9 9 2 V 9 10 6 10 (0.6). . . . = 150 V

ELECTROSTATICS www.physicsashok.in 29 Example-39 : Ashort electric dipole is situated at the origin of coordinate axis with its axis along x-axis and equator along y-axis. It is found that the magnitudes of the electric intensity and electric potential due to the dipole are equal at a point distance r = 5 m fromorigin find the position vector of the point. Sol. Let P be such a point at distance r and angle . fromequator. Now |EP| = |VP| or P 2 P 3 2 K 1 3sin K sin r r . . . . or 1 3sin2 sin 5 . . . . Squaring both nets y x rP q 1 + 3 sin2. = 5 sin2. or sin 12 . . or ..= 45º Positive vector of r . point P is r 5 (�i �j) 2 = + . Electric Dipole in Uniform Electric Field : (i) Torque :When a dipole is placed in a uniformelectric field as shown in Fig. the net force on it, F . ..qE . ..q. E.. . 0 .. .. while the torque, �q +q P qE qE l E . = qE × 2l sin . . . pEsin . . or . . p . E . . ..

Fromthis expression it is clear that torque acting on a dipole ismaximum(= pE)when dipole is perpendicular to the field andminimum(=0) when dipole is parallel or antiparallel to the field. NOTE :If the electric field is not uniform the dipole will experience both a resultant force and a torque so its motion will be combined translatory and rotatory (if .. 0 or 180º) (ii) Work :Work done in rotating a dipole in a uniformfield througha small angle d.willbe, dW= .d..= pE sin .d. So,work done in rotating a dipole fromangular position .1 to .2with respect to field, 2 1 1 2 W pEsin d pE [cos cos ] . . . . . . . . . . So, if a dipole is rotated fromfield direction, i.e., .1 = 0 to position ., i.e., .2 = .. W= pE[1 � cos.]

ELECTROSTATICS www.physicsashok.in 30 �q +q P= 0 E Wmin = 0 (a) +q �q P = 180º E (b) Wmax = 2pE (iii) Potential energy : In case of a dipole (in a uniformfield), potential energy of dipole is defined aswork done in rotating a dipole froma directionperpendicular to the field to the given direction. i.e., U = (W. � W90º) U = pE (1 � cos .) � pE (1 � cos 90º) U . .pE cos. . .p . E . .. (iv) AngularSHM:Adipolewhen placed in a uniformelectric field, align itself parallel to the field. If it is given a small angular displacement . about its equilibriumposition, the restoring torque produced in itwillbe, . = �pE sin . = � pE. (. sin . . .) or 22 d I pE dt. . . . or 2 2 2 ddt. . .. . with 2 pEI . . This is standard equation of angular SHMwith period T 2. .. . . . . . . . So, dipolewill execute angular SHMwith time period T 2 I PE . . Interaction of Two Dipoles : If a dipole is placed in the field ofother dipole, then depending onthe positions of dipoles relative to eachother, force, torque and potential energyare different. TABLE : Dipole-Dipole Interaction S.No. Relative position of Dipole Force F Torque . PotentialEnergyU 1. r

p1 F F p2 1 2 0 4 1 6p p (along r) 4 r . .. zero 1 2 3 0 1 2p p 4 r . .. 2. p r 1 p2F F 1 2 4 0 1 3p p 4.. r (along r) zero 1 2 3 0 1 p p 4.. r 3. r F p F p 2 1 1 2 4 0 1 3p p 4 r . .. on 1 2 1 3 0 p , 1 2p p 4.. r zero (perpendicular to r) on 1 2 2 3 0 p , 1 p p 4.. r Example-40 : Two point charges +3.2 ×10�19 C and �3.2 × 10�19 C are placed at a distance 2.4 × 10�10m, from each other. This dipole is placed in a uniformelectric field of 4.0 × 105V/m. (a) Find the necessarytorque required to rotate the dipole through 180º fromits equilibriumposition. (b) What is thework done inrotating the dipole through 180º ?

ELECTROSTATICS www.physicsashok.in 31 (c) What is the potential energy of dipole in this position ? Sol. Electric dipolemoment p = q × 2l Here, q = 3.2 × 10�19 C 2l = 2.4 × 10�10 m . p = (3.2 × 10�19) (2.4 × 10�10) p = 7.68 × 10�29 C-m (a) necessary torque . = pE sin . ..= 7.68 × 10�29 × 4 × 105 × sin 180º ..= 0 ( . sin 180º = 0) (b) Work done inrotating the dipole through 180º is given by W= pE (cos.1 � cos.2) Here, .1 = 0, .2 = 180º, p = 7.68 × 10�29 C-m and E = 4.0 × 105 V/m . W = 7.68 × 10�29 × 4 × 105 × (cos 0º � cos 180º) W = 7.68 × 10�29 × 4 × 105 × (1 + 1) W = 6.14 × 10�23 J (c) The potential energyof dipole in this positionis given by U. = � pE cos . where ..is the angle between the axis of dipole and electric field. In equilibriumposition, ..=0 . U0 = � pE or U0 = � (7.68 × 10�29) (4.0 × 105) U0 = �3.07 × 10�23 J Example-41 : Awheelhavingmassmhas charges +q and �q on diametrically opposite points. It remains in equilibriumon a rough inclined plane in the presence of uniform vertical electric field E = +q �q E (A) mg q (B) mg 2q (C) mg tan 2q q (D) none Sol. The torque of electric force about centre is balanced bytorque due to friction about the centre. r f= PE sin. . . = PE sin. But f=mg sin. (for equilibrium) . mgr sin. =PE sin. +q -q mgsin 90- 90- E f P E mgr mgr P q 2r

. .

.

. P = q × 2r (dipolemoment) E mg 2q .

ELECTROSTATICS www.physicsashok.in 32 Example-42 : Anonconducting ring ofmassmand radius Ris charged as shown. The charged densityi.e. charge per unit length is .. It is then placed on a rough nonconducting horizontalsurface plane.At time t = 0, a uniformelectric field E = E0i .. is swithced on and the ring start rollingwithout sliding. y � x � � � +++ + Determine the frictionforce (magnitude and direction) acting on the ring, whenit startsmoving. Sol. Where f is friction surface. f = macm ...(1) Let d. element on ring then dq . . d. dq . .Rd. { d dR . . . . } y � x � � � +++ + p f rough non-conducting surface d dP E . . . ... .. d. . 2RdqEsin .. .. . {. dp = (2Rdq)} d. . 2RE .Rsin.d. / 2 2 E 0 . 2R E . sin d . . . . . 2 E . . 2R .E 2R2 .E � fR = mR2. �dq f dqq q dq E d p p- q 2R.E � f = mR.

2R.E � macm = mR. (fromeqn (1)) for pure rolling, acm = R. . 2R.E = macm + mR. 2R.E = 2 macm macm = .RE cm 0 f . ma . .RE Example-43 : In the figure shown S is a large nonconducting sheet of uniformcharge density ..Arod Rof length l andmass �m�is parallel to the sheet and hinged at itsmid point. The linear charge densities on the upper and lower half of the rod are shown in the x l - l R Ss figure. Find the angular acceleration of the rod just after it is released. Sol. Electric field due to non-conducting sheet is 0 . 2 . . d. = x E dq × 2 = 2x Edq / 2 2 /2 0 0 d 2x E dx 2E x2 . . . . . . . . . . . . . . . . . 2E 2 E 2 8 4 . . . . . . . . E 2 I 4 . . . . Edq Ss Edq �dq +dq xx or m 2 E 2 12 4 . . . . .

ELECTROSTATICS www.physicsashok.in 33 . 2 2 E 12 3E 4 m m . . . . . . . . . 0 3 2 m . .. . . ELECTRIC FLUX Electric flux throughan elementaryarea dS is defined as the scalar product of area and field, i.e., d.E . E . dS .. . d.E = EdS cos . n E i.e., E dS . . . E . dS .. . It represents the total lines offorce passing through the given area. Regarding electric flux following points areworthnoting : (a) Electric fluxwillbemaximumwhen cos. = 1, i.e., . = 0, i.e., electric field is normal to the area with (d.E)max = EdS. En (d ) = EdS E max (b) Electric fluxwillbeminimumwhen cos. = 0, i.e., . = 90º, i.e., field is parallel to the surfacewith (d.E)min = 0. E n (d ) = 0 E min (c) For a closed body outward flux is taken to be positive [Fig. (a)] while inward negative [Fig. (b)] n (a) Positive flux (b) Negative flux GAUSS�S LAW This lawgives a relation between the net electric flux through a closed surface and the charge enclosed by the surface. According to this law, �the net electric flux through any closed surface is equal to the net charge inside the surface divided by .0.� In symbols it can bewritten as, in E S 0 . . E . n� dS . q. . .. . ...(i)

but this formofGauss�s lawis applicable only under following two conditions : (i) The electric field at every point onthe surface is either perpendicular or tangential. (ii) Magnitude ofelectric field at everypoint where it is perpendicular to the surface has a constant value (sayE). Here, S is the areawhere electric field is perpendicular to the surface. Applications of Gauss�s Law : To calculate Ewe choose animaginaryclosed surface (calledGaussian surface) inwhichEqs. (i) or (ii) can be applied easily. Inmost of the caseswewill use Eq. (ii). Let us discuss fewsimple cases.

ELECTROSTATICS www.physicsashok.in 34 (i) Electric field due to a point charge : FromEq. (ii), in 0 ES . q. Here, S = are of sphere = 4.r2 and qin = charge enclosing theGaussian surface = q q r E . . 2 . 0 E 4.r . q . . 2 0 E 1 q 4 r . .. It is nothing but Coulomb�s law. (ii) Electric field due to a linear charge distribution : Consider a long line chargewith a linear charge density(charge per unit length) ..We have to calculate the electric field at a point, a distance r fromthe line charge.We construct aGaussian surface, a cylinder of any arbitrary length l of radius r and its axis coinciding with the axis of the l r E ++++ + + + + E line charge. Hence, we can apply theGauss�s lawas, in 0 ES . q. E Curved Surface E Plane Surface Here, S = area of curved surface = (2.rl) and qin = net charge enclosing this cylinder = .l . 0 E(2 ) . .. . . l l . 0 E 2 r . .

.. i.e., E 1r . (iii) Electric field due to a plane sheet of charge : FromGauss�s Law in 0 ES . q. . 0 0 0 E(2S ) ( ) (S ) . . . + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + E E S0 . 0 E 2. . . (iv) Electric field near a charged conducting surface : This is similar to the previous one the onlydifference is that this time charges are on bothsides.Hence, applying in 0 ES . q.

ELECTROSTATICS www.physicsashok.in 35 Here, S = 2S0 and qin = (.) (2S0) + E S0 E + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + . + 0 0 0 E(2S ) ( ) (2S ) . . . . 0 E . . . Example-44 : Which of the following statements is/are correct ? (A) Electric field calculated byGauss lawis the field due to only those charges, which are enclosed inside the Gaussian surface. (B) Gauss law is applicable onlywhen there is symmetrical distribution of charge. (C) Electric flux through a closed surface is equal to total flux due to all the charges enclosed within that surface only. (D) None of these Sol. Since, electric field at a point is equal to electric flux passing per unit area, therefore, . E. dS .. . . is the net flux emanating from a closed surface. Though net flux through the closed surface depends upon the charges enclosed in that surface only but electric field E at a point depends not onlyupon charges enclosed but it depends upon charges lying outside the surface also. Hence (A) is wrong. Gauss lawis applicable to a closed surface. The surfacemay have any shape. It means, it is a general law. Hence (B) is wrong. Gauss law is 0E. dS q . . . . . .. . . It means, net flux through a closed surface depends upon .q. But it is equal to net charge enclosed within the surface only. Hence (C) is correct. Obviously (D) is wrong.

Example-45 : In a region of uniformelectric field E, a hemispherical body is placed in such away that field is parallel to its base (as shown in figure). E OC The flux linked with the curved surface is : (A) zero (B) �.R2E (C) .R2E (D) R2 E 2 . Sol. The flux linked with the body is zero as it does not enclose any charge. . = .cs + .pb = 0 As field is parallel to base, the flux linked with base .pb = E × .R2 cos 90º = 0 E n . .paralled to base = 0 .curved surface = 0 Example-46 : If a point charge q is placed at the centre of a cubewhat is the flux linked (a) with the cube (b) with each face of the cube ? Sol. (a)According to Gauss�s Law, in total 0 0 . . q . q . . (b) Nowas cube is symmetrical bodywith 6-faces and the point charge is at its centre, so electric flux linkedwith each facewill be total 0 ´ q 6 6 . . . . .

ELECTROSTATICS www.physicsashok.in 36 NOTE : If charge is not at the centre of cube (but any where inside it), total flux will not change but the flux linked with different faces will be different. Example-47 : If a point charge q is placed at one corner of a cube, what is the flux linkedwith the cube ? Sol. By placing three cubes at three sides of given cube and four cubes above, the chargewill be in the centre. So, the fluxwith each facewillbe one-eight of the flux (q/.0). . Flux throughthe cube 0 q 8 . . Example-48 : Calculate flux through the cube and flux through the each surface of cube when q is placed at one of its corner. Sol. To cover charged particle completlywe have to use 7 extra cubes around the charged particle. So, Flux Through cube 0 ( ) q 8 . . . Flux through 0 ABCD q 3 24 . . . . Flux through 0 ABFE q 3 24 . . . . D C AE F B HG q Flux through 0 ADEH q 3 24 . . . . Flux through EFGH = 0

Flux through FGCB = 0 Flux through GCDH = 0 Example-49 : The electric field in a region is given by 0 E = E x i .. . l . Find the charge contained inside a cubical volume bounded by the surfaces x = 0, x = a, y = 0, y = a, z = 0 and z = a. Take E0 = 5 × 103N/C, l = 2cm and a = 1cm. Sol. 0 E . E x �i .. . Totalflux 0 . q . 0 EA . q . whereAis area of surface 0 2 0 E a . a . q . . . 3 0 0 E a q . . . 3 0 0 E a q . . . . q . .3 12 2 3 2 5 10 8.85 10 1 10 2 10. . . . . . . . . . q 5 103 8.85 10 4 2 . . . . . . q.2.2.10.12 C

ELECTROSTATICS www.physicsashok.in 37 ELECTRIC FIELD AND POTENTIAL DUE TO CHARGED SPHERICAL SHELL OR SOLID CONDUCTING SPHERE Electric Field The allpoints inside the charged spherical conductor or hollowsphericalshell, electric field E . 0, .. as there is no charge inside such a sphere.We can construct a Gaussian surface (a sphere) of radius r > R. FromGauss�s Law C 0 E . ds . q . . .. . 2 0 E(4.r ) . q . . 2 0 E 1 q 4 r . .. r R Gaussian surface E + q +++ + + + + + ++ + Hence, the electric field at anyexternalpoint is the same as if the total charge is concentracted at centre. At the surface of sphere r = R, . 2 0 E 1 q 4 R . .. Thus, we canwrite, Einside = 0 surface 2 0 E 1 q 4 R . .. outside 2 0 E 1 q 4 r .

.. The variation ofelectric field (E) with the distance fromthe E O R r 1 qR2 = E r2 1 centre (r) is as shown in fig. NOTE : � At the surface graph is discontinuous. � 2 surface 2 0 0 0 E = 1 q = q/4pR = s 4pe R e e Potential : As we have seen, outside 2 0 E 1 q 4 r . .. . outside 2 0 dV 1 q dr 4 r . . . . . . . . .. E dV dr . . . . .. .. . V r 0 outside 2 0 dV q dr 4 . r . . .. . . (V. = 0) . 0 V 1 q or V 1 4 r r . . .. Thus, at externalpoints the potentialat anypoint isthe samewhenthewhole charge is assumedto the concentrated at the centre.At the surface of the sphere, r = R. . 0 V 1 q 4 R . .. At some internalpoint electric field is zero everywhere, therefore, the potential is same at allpointswhich is equal to the potential at surface. Thus,we canwrite,

inside surface 0 V V 1 q 4 R . . ..

ELECTROSTATICS www.physicsashok.in 38 and outside 0 V 1 q 4 r . .. V O R r 1 qR = V r 1R The potential (V) varieswith the distance fromthe centre (r) as shown in Fig. Example-50 : The diameter of hollowmetallic sphere is 60 cmand charge on sphere is 500 µC. Find the electric field and potential at a distance 10 cmfromcentre of sphere. Sol. The point at 10 cmdistance fromcentre of spherewillbe inside the sphere. Inside hollowconducting sphere, electric field is zero everywhere and potential is uniform(same as at the surface). Hence, intensityof electric field E = 0 and potential 0 V 1 q 4 R . .. Here, q = 500 µC = 500 × 10�6 C and R = 30 cm= 0.30 m . . . . . 6 9 500 10 V 9.0 10 0.30 . . . . . V = 1.5 × 107 V ELECTRIC FIELD AND POTENTIAL DUE TO A UNIFORMALY NON-CONDUCTING SPHERE Electric field : Let positive charge q is uniformlydistributed throughout the volume of a solid sphere of radius R.We have to find the intensityof electric field due to this charged sphere at point P distance r fromcentreO. ApplyingGauss�s law in 0 ES . q. ...(i) Here, S = 4.r2 and 3 in q ( ) 4 r 3 . . . . . .. .. Here, 3 . = charge per unit volume = q 4/3.R

Substituting these values inEq. (i),we have 3 0 E 1 . q . r 4 R . .. or E . r At the centre r = 0, so, E = 0 At surface r = R, so, 2 0 E 1 q 4 R . .. To find the electric field outside the charged sphere,we use a sphericalGaussian surface ofradius r(>R).This surface encloses the entire charged sphere, so qin = q, and Gauss�s law gives. 2 0 E(4.r ) . q . or 2 0 E 1 q 4 r . .. or 2 E 1 r . Thus, for a uniformly charged solid sphere,we have the following formulae formagnitude of electric field. inside 3 0 E 1 . q . r 4 R . .. surface 2 0 E 1 . q 4 R . .. outside 2 0 E 1 . q 4 r . .. E O R r 1 qR E r E r2 2 1

ELECTROSTATICS www.physicsashok.in 39 The variation of electric field (E)with the distance fromthe centre of the sphere (r) is shownin fig. Potential : The field intensityoutside the sphere is, outside 2 0 E 1 . q 4 r . .. outside outside dV E dr . . . V r outside 2 0 dV 1 . q dr . . 4 r . . .. . . . 0 V 1 q as V 0 4 r . . . .. or V 1r . At r = R, 0 V 1 q 4 R . .. i.e., at the surface of the sphere potential is S 0 V 1 q . 4 R . .. The electric intensityinside the sphere, inside 3 0 E 1 . q . r 4 R . .. inside inside dV E dr . . .

S V r V inside 3 R 0 dV 1 q r dr 4 R . . .. . . . 2 r S 3 0 R V V 1 q r 4 R 2 . . . . . . . .. . . Substituting S 0 V 1 . q , 4 R . .. we get22 0 V 1 . q 3 1 r 4 R 2 2 R . . . . . . .. . . At the centre r = 0 and C S, 0 V 3 1 q 3 V 2 4 R 2 . . . . . . . .. . i.e., potential at the centre is 1.5 times the potential at surface. Thus, for a uniformlycharged solid spherewe have the following formulae for potential. outside 0 V 1 q 4 r . .. surface 0 V 1 q 4 R . .. and 2 inside 2 0 V 1 q 3 1 r 4 R 2 2 R . . . . . . .. . . 1 q

R 1 qR 32 V The variationof potential (V)withdistance formthe centre (r) is as shown in figure.

ELECTROSTATICS www.physicsashok.in 40 Example-51 : The radius of a solidmetallic non-conducting sphere is 60 cmand charge on the sphere is 500 µC. Find the electric field and potential at a distance 10 cmfromcentre of sphere. Sol. The point 10 cmfromcentre of spherewill be inside the sphere. Hence, inside 3 0 E 1 . qr 4 R . .. Here, q = 500 × 10�6 C, r = 10 cm = 0.10 m R = 60 cm= 0.6 m . . . 6 9 6 inside 3 E 9.0 10 500 10 0.10 2 10 N/C (0.60) . . . . . . . . and 2 inside 2 0 V 1 q 3 1 r 4 R 2 2 R . . . . . . .. . . . . 6 2 9 2 9 10 (500 10 ) 3 1 (0.10) (0.60) 2 2 (0.60) . . . . . . . . . . . . 7.5 106 3 1 2 72 . . . . . . .. .. 7.5 106 107 72 . . . = 1.1 × 107 V Example-52 : Asolid non conducting sphere ofradiusRhas a non-uniformcharge distribution ofvolume charge density, 0 r , R . . . where .0 is a constant and r is the distance fromthe centre of the sphere. Showthat: (a) the total charge on the sphere isQ = ..0R3 and (b) the electric field inside the sphere has amagnitude given by, 2 4 KQr E . R .

Sol. (a) dv = 4.r2dr 0 2 dq dV r 4 r dr R . . . . .0 3 4 r dr R . .. R 0 3 0 4 Q r dr R . . .. r dr R 4 0 3 0 4 R Q Q R R 4 . . . . .. .. (b) r 0 3 in 0 4 q r dr R . . .. 4 4 0 0 4 r 4 r R 4 4R . . . .. .. in 2 0 E q 4 r . . . 4 0 2 0 4 r E 4R 4 r . . . ...

But Q = ..0R3

ELECTROSTATICS www.physicsashok.in 41 0 3 Q R .. . and 0 K 1 4 . . . . 2 4 E KQr R . Example-53 : Three concentric metallic shellsA, B and C of radii a, b and c (a < b < c) have surface charge densities ., �. and . respectively. (i) Find the potential of three shellsA, B and C. (ii) If the shellsAand C are at the same potential, obtain the relation between the radii a, b and c. Sol. The three shells are shown in figure. (i) Potential ofA= (Potential ofAdue to + ..onA) + (Potential ofAdue to � ..on B) + (Potential ofAdue to + ..on C) 4 a2 4 b2 4 c2 K a b c . . . . . . . . . . . . . . . . . 0 . a . b . c .. c b C B A a +s +s - s Potential of B = (Potential due to + ..onA) + (Potential due to � ..on B) + (Potential due to + ..on C) 4 a2 4 b2 4 c2 K a b c . . . . . . . . . . . . . . . 2 0 a b c b . . . . . . . . . . .

Potential of C = (Potential due to + ..onA) + (Potential due to � ..on B) + (Potential due to + ..on C) 4 a2 4 b2 4 c2 K a b c . . . . . . . . . . . . . . . 2 2 0 a b c b c . . . . . . . . . . . (ii) Given that VA =VC or . . 2 2 0 0 a b c a b c b c . . . . . . . . . . . . . . . . or a2 b2 a b c c b c . . . . . Solving we get, c = (a + b). Example-54 : An electric field converges at the originwhosemagnitude is givenby the expression E = 100N/C, where r is the distancemeasured fromthe origin. (A) total charge contained in any sphericalvolumewith its centre at origin is negative. (B) total charge contained at anysphericalvolume, irrespective ofthe location of its centre, is negative. (C) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude 3×10�13 C. (D) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude 3×10�9 C.

ELECTROSTATICS www.physicsashok.in 42 Sol. Since, electric lines of forces are terminated into the sphere. Hence, total charge contained by should the sphere should be negative. So, options (A) and (B) are correct E E E E . 2 0 E q 4 r . .. Here r = 3 × 10�2C . q = 3 × 10�13 m . Hence, option (C) is correct. Example-55 : Figure shows a section through two long thinconcentric cylinders of radii a&bwith a < b. The cylinders have equal and opposite charges per unit length .. Find the electric field at a distance r fromthe axis for b a (A) r < a (B) a < r < b (C) r > b Sol. a l b - l (A) r < a C 0 0 E . ds . . . .. . r b Ea 0 E . 2 r . O. . . E = 0 (B) a < r < b C 0 q E . ds . . . .. . r b a 0 E . 2 r . .. . . . 0 E 2 r . . .

. (C) r > b C 0 E ds 0 . . . . . . .. . . . . . E . 2.r. . 0 a b r E . 0 THINKINGTYPEPROBLEMS 1. The electric charge ofmacroscopic bodies is actually a surplus or deficit of electrons.Why not protons? 2. What are insulators and conductors? 3. Acharged rod attracts bits of dry cork dust which, after touching the rod, often jump away fromit violently. Explain. 4. Atruck carrying explosives has ametal chaing touching the ground.Why? 5. Howis the Colulomb force between two charges affected bythe presence of a third charge ? 6. Aperson standing on an insulating stool touches a charged insulated conductor. Is the conductor completely discharged ? 7. Electric lines of fore never cross.Why ?

ELECTROSTATICS ELECTROSTATICS 8. The electric field inside a hollow charged conductor is zero. Is this true or false ? 9. The electric field inside a hollow charged sphericalconductor is the same at allpoints and is equalto the field at the surface. Is this true or false ? 10. Theelectricpotentialinsideahollowchargedsphericalconductor isthesameat allpointsandisequalto the potential of the surface. Is this true or false ? 11. AchargeAsituatedoutsideanunchargedhollowconductorexperiencesaforceifanother chargeBisplaced insidetheconductor, but Bdoesnot experienceanyforce.Why?Doesitnotviolatethethirdlawofmotion? 12. If only one charge is available, can it be used to obtain a charge many times greater than it in magnitude? 13. In the previous question, does it make a difference which face of B is touched in order to remove the free charge (include positive charge) from B? 14. Can a small spherical body of radius 1 cm have a static charge of 1 C? 15. A metal leaf is attached to the internal wall of an electrometer insulated from the earth. The rod and housing of the electrometer are connected by a wire, and then a certain charge is imparted to the housing. Will the leaves of the electrometer be deflected ? What will happen to the leaves if the wire is removed and the rod is then charged ? 16. Abroad metalplate is connected to the earth through a galvanometer. Apositively charged ball flies along a straight line above the plate at a distance much less than the linear dimensions of the plate. Draw an approximate diagram showing how the current flowing through the galvanometer depends on time. 17. The work done in carrying a point charge from one point to another in an electrostatic field depends on the path along which it is taken. Is this true of false? Briefly explain your answer. 18. The figure shows lines of constant potential in a region in which on electric field is present. The values of the ptentials are written in brackets. Of the points A, B and C, the magnitude of the electric field is greatest at the point...... Give reasons for your answer in brief. 19. Can two balls with like charges be attracted to each other? 20. The housing of the electrometer of question no.15 is given a charge. Will the leaves of the electrometer be deflected in this case? Will the deflection change if the rod is earthed ?There is no connecion between the rod and the housing. 21. By touching different points of a metal bucket with a test ball B connected by a wire to an earthed electrometer it can be observed that deflection ofthe leaves of the electrometer is the same for any position of the ball. But if the wire is removed and the charge is transferred by the ball to the ball of the electrometer, the deflection depends on which surface of the bucket (external or internal) is touched before that. Why ?

22. The gravitationalfield strength is zero inside a sphericalshellof matter. The electric field strength is zero not only inside anisolated charged spherical coonductor but inside an isolated conductor of anyshape. Is the gravitational fieldinside,say, acubicalshellofmatter zero ?Ifnot, inwhatrespect istheanalogynot complete?

23. Cantwo equipotentialsurfacesintersect? 24. An isolated, conducting spherical shell of radius R carries a negative charge Q. What will happen if a small, positivelychargedmetalobject isplacedinsideandconnectedbyawire?Assume that thepositivechargeis (a) less than, (b) equal to and (c) greater than Q. 25. AnunchargedmetalspheresuspendedbyasilkthreadisplacedinauniformexternalelectricfieldE.What is the magnitude of the electric field for points inside the sphere ? Will your answer change if the sphere carries a charge? 26. Is charge uniformlydistributed over the surface of an insulated conductor of anyshape ? If not, what is the rule for the distribution of charge over a conductor? 27. Areequipotentialsurfaceswhichariseduetoapointchargeandwhosepotentialsdiffer byaconstant amount (say 1 volt) evenly spaced in radius ? 28. Two point charges Q and 4Q are fixed at a distance of 12cm from each other. Sketch the lines of force and www.physicsashok.in 43

ELECTROSTATICS www.physicsashok.in 44 locate the neutralpoint, if any. 29. Acharge +Qis fixed at a distance d in front of an infinite metal plate. Draw the lines of force indicating the direction clearly. 30. Indicate the surface distribution of charge on a squaremetal plate byusing dots in such away that the greater the surface densityof charge, the farther awaythe dots are fromthe plate (shown in fig.). 31. Ametal plate ismoved with a constant acceleration a parallel to its plane. Does any charge develop on its surfaceswhich are perpendicular to itsmotion? 32. Is an electric field of the type shown by the electric lines in the figure physicallypossible? 33. Zero work is donewhen a charged particle is transferred fromone equipotential surface to another. Is this true or false ? 34. ApotentialdifferenceVis set up betweena filament emitting electrons in avaccumtube and a thinmetallic ring. The electron beampasses past the ring through its central regionwithout spreading. The kinetic energyof the electrons inthe beamincreaseswhile the batteryproducing the potentialdifferenceVperforms nowork, since no current flows through the circuit. Howcan his fact be reconciledwith the principle of conservation of energy? 35. There is an electric field near the surface of a conductor carrying direct current. Is this true or false? 36. Ordinary rubber is an insulator. But the special rubber tyres of aircrafts are made conducting.Why is this necessary? 37. Asmall sphere is charged to a potentialof 50Vand a big hollowsphere is charged to a potentialof 100V. How can youmake electricityflowfromthe smaller sphere to the bigger one? 38. An electric dipole is placed in a non-uniformelectric field. Is there a net force on it? 39. Apoint charge is placed at the centre of a spherical gaussion surface. Does electric flux e . change (a) if the sphere is replaced by a cube of the same volume, (b) if a second charge is placed near, and outside, the original sphere, and (c) if a second charge is placed inside the gaussion surface? 40. Aspherical rubber balloon carries a charge that is uniformlydistributed over its surface.As the balloon is blown up, how does E vary for points (a) inside the balloon, (b) on the surface of the balloon, and (c) outside the balloon? THINKINGPROBLEMS SOLUTION 1. Because protons are tightlybound in the nucleus.They cnnot be removed fromthere easily. 2. Conductors arematerials inwhichthere are a fewfree electrons per atomof thematter. Insulators arematerials inwhich the electrons are not as free as in conductors. 3. Acharged rod first attracts the dust by producing unlike charge at the near end and like charge at the far end. When the cork dust touches the rod, however, it acquires like charge and so is repelled strongly bythe charge on the rod. 4. To conduct away the charge produced by friction. This charge, unless conducte

d away to the earth, may produce sparks causing the explosives in the truck to catch fire and explode. 5. The force on the charge due to another does not depend on the presence of other charges, that is, electrical forces are physically independent. 6. No, theman acts as a conducting body. He shares chargewith the charged body. 7. Because if theycross, the electric field at the point ofintersectionwillhave two directions simultaneously,which is physicallyimpossible. 8. It is true. The electric field inside a hollowcharged conductor is zero. This follows fromthe conditionthat all points on conductor have the same potential.

ELECTROSTATICS www.physicsashok.in 45 9. It is false. The field inside is zero but on the surface, the field is finite. 10. It is true. 11. This is so because there is an electric field overAbut there is no electric field over B as it lies inside a hollow conductor.No, this is not in violationof the third lawofmotion. In fact, force arises betweenAand the hollow sphere, the charge B is simplyan internal part of the sphere. 12. Yes, by repeating the induction process. Place an insulated conductor B close to the charged body.A. yinduction, an equal unlike charge is induced at the near end ofBand equal like charge at the far end.Touch the body B with your finger tip in the presence ofA.The like charge at the far end will flowto the earth. But the unlike charge at the near endwill remain bound due to the force of attraction. Nowremove the bodyB to a distant place. The unlike charge, equal inmagnitude to the inducing charge,will spread over the surface ofB as there is no force of attraction to keep it on one side. Next, deliver this charge to a big body by conduction, that is, by touching this bodywithB. Repeat the process after completely dischargingB. To obtain a similar charge in largemagnitude, first deliver unlike charge to a bodyC and thenrepeat the induction process betweenC and the body inwhich like charge is to be stored. 13. No. The free chargewillgo to the earth via the finger, irrespective ofwhich part ofB is touched, because the free positive charge inB exists at a higher potential due to the positive charge onA. It is a fact that positive charge always flows fromhigher to lower potential, irrespective ofthe path. 14. E 9 109 1 9 1013 NC 1 0.01 . . . . . .The field is everylarge.Air cannot sustain sucha strong electric field, so a small bodycannot have a charge of 1 coulomb. 15. The housing and the rod connected togetherwillhave the same potential, so the leaveswillnot be deflected. After thewire is removed and the rod is charged, both the leaves will be deflected because of the potential difference between the rod and the housing. 16. As the charge approaches the plate, electrostatic induction causes the induced positive charge to pass into the earth,while the induced negative charge accumulates on the upper surface of the plate.Apositive current pulse passes through the galvanometer. No current is produced when the chargemoves above the plate.An opposite current is producedwhen the chargemoves away from the plate. 17. It is false. thework done in carrying a point charge fromone point to another does not depend on the path alongwhich it is carried because electric field is conservative. 18. It is greatest at the pointB. Since the electric field is the rate of fallof potential, the stronger the field, the closer the equipotentialsurfaces. In the figure, the equipotential surfaces are closest in the neighbourhood ofB, so the field at Bis the greatest. 19. Theycan, if the charge of one ball ismuch greater than that of the other.The forces of attraction caused by the induced chargesmay exceed the forces of repulsion. 20. When the housing of the electrometer is given some charge, the potentialof the rod cannot be the same as that

of the interior of the housing because the rod lies partlyoutside.On account of thepotentialdifference between the rod and the housing, the leaveswill be deflected. When the rod is earthed, the potentialdifference between the rod and the housing is increased further, so the deflectionwillbe greater. 21. The electrometermeasures the potentialdiference between the bodyand the housing of the electrometer. Since the surface of the bucket is equipotential, the leaves showthe same deflectionwherever the testing ballBmaytouch the surface of the bucket. But when the ballB is disconnected fromA, the ballBcollects charge by conduction.When it delivers this charge toA, the potentialof the leaves (and also of the rod) is raised above the potential of the housing, so there is deflection of the leaves.When B touches the inner

ELECTROSTATICS www.physicsashok.in 46 surface, it collects no charge because there is no potential difference between it and the inner surface of the bucket. Unless there is a potentialdifference, there can be no flowof charge fromone body to another. So no deflection takes placewhen the ballB touches the inner surface and then touchesA. But when it touches the outer surface, it collects a certain amount of charge, so there is deflection. 22. No, the field inside a cubical shellofmatter is not zero. The analozyis not complete in respect of distribution of mass and charge.Mass is uniformlydistributed but charge is not distributed uniformly. It ismore concentrated at the edges and corners. 23. No, two equipotentialsurfaces can never intersect because if they intersect at a point, the electric field at that point can have two directions simultaneously,whichis physicallyimpossible. 24. For the sake of simplicity, let us consider themetal bodyto be a sphere placed at the centre of the shell. VA (potentialof the surface of the shell) 0 1 Q q 4 R . . . .. where R= radius of the shell and q is the charge of themetal body. VB (potential of themetal body) 0 1 q Q 4 r R .. . .. . . . . . .... . . B A 0 V V q 1 1 4 r R . ... . .. . .. .... . . B A V . V So chargewillflowfromthemetalbodyto the shell. SinceVB =VA at equilibrium, qmust be reduced to zero, i.e., the entire charge on themetalbodymust flowto the outer shell. V, the common potential 0 q Q 4 R . . .. . Thus, (a) when q <Q,Vis negative, (b) when q =Q,V= 0, (c) when q > Q, Vis positive. 25. The field inside themetal sphere at all points is zero because r . is infinityformetals and 0 r E , E . . So E = 0. No, the answer does not change if the sphere carries a charge.

26. Charge is not distributed uniformlyover the surface ofa conductor ofanyshape. Charge is distributed according to the curvature of the surface.The greater the curvature, the greater the surface density of charge. 27. No. . . 2 1 1 2 0 1 2 0 1 2 q 1 1 q r r V V 4 r r 4 r r .. .. . . . .. . ... .. .. .. .. or . . 0 1 1 1 q r 4 r r r . . .. . . where 2 1 .r . r .r Obviously, . r is not constant. So the equipotential surfaces differing by 1Vare not equispaced. 28. The lines of force are shown in the figure. Here, all the lines near the edge of the figurewill appear to radiate uniformlyfromthe point P, the centre ofgravity of the charges.At the neutral point N, the total field is zero. Let it be at a distance x from4Q. Then . .2 2 0 0 1 4Q 1 Q 4 x 4 12 x . .. .. . or x = 8 cm 29. The lines of force are shown in the figure. Explanation Since it is ametalplate, its surface is an equipotential surface. So lines of forcemust terminate normallyon it. Note : One everyimportant result follows fromthismap of electric flued due to a point charge and the induced charges on ametalplate. If a charge �Qis placed as far behind the plate as +Q

ELECTROSTATICS www.physicsashok.in 47 is infront and the plate is removed,AB(positionof the plate) is still an equipotential surface. So for interaction between +Qand the plate or for electric fields between them,wemayuse the formalismof replacing the plate by �Q. This is called the electrical image of +Q. 30. The distributionof charge is shownin the figure. Explanation Since the surface density of charge is proportional to curvature and corners have the greatest curvature and a plane surface has zero curvature, dots are equidistant fromthe straight portion and far away fromthe corners. 31. Since the plate is accelerated, the electrons inside are also acceleratedwith the same acceleration.Aforcemay be applied on an electron, onlywhen an electric field is created. So theremust develop charges on the faces that are perpendicular to the direction ofmotion, so that there develops an electric field inside themetal plate. The front facewill be charged positively, and the rear one negatively.The intensityof the electric field is given by eE = ma. Since E = ./.0, the surface density of charge developed is given by . = .0E = .0ma/e 32. No. The concentration of lines at the botomindicates a field which is stronger at the botoomthan on top. Imagine a rectangular pathwith its two sides perpendicular to the electric lines.Nowmove a charge along this path. Some networkwillbe performed.But inanelectric field,work done is essentiallyzero as it is a conservative field. Hence, an electric field of the type represented is physically impossible. 33. False,Work is always done byanexternal agent when a charged particle is transferred fromone equipotential surface to another. Thework done per unit charge = potential difference. 34. There is an electric field between the filament and the ring.As the electron is emitted fromthe filament, it experiences a force towards the ring and so it is accelerated. It electricalpotential energy,which it acquires on emerging fromthe filament, is converted into its kinetic energy. Thus, though the batterydoes not supply any energy, the electron can gain velocityat the cost of its ownpotential energy. 35. True. There is an electric field inside a current-carrying conductor.This is equal to V/l, whereVis the voltage across the conductor and l is the length of the conductor. Now, consider a closed path abcda as shown in the figure.Work done byan external agent along abcda is zero. . inside outside inside E .ab.0.E .cd.0.0 . E . Eoutside Hence, there is a field near the surface of the conductor. 36. To conduct awayelectricityproduced byfriction. 37. Place the smaller one inside the bigger one.As potential inside a hollowconductor is the same as that of its surface, the potential inside the hollow conductor is 100V. The potential of the smaller conductor is now 100+50 =150V. Connecte the two conductors byawire. Chargewill flowfromthe smaller one to the bigger one as the smaller one is at 150Vand the bigger one at 100V. 38. Yes, there is a net force on the dipole given by F p E .

. .

.l 39. Total flux E 0 . . E ..s. 1 q . . .. . , byGauss�s law. (a) No, the fluxwill not change as it depends onlyon the total charge inside the surface and not on the extent and shape of the surface. (b)No, it remains the same as the total fluxis determined solely bythe charge inside the surface and not the charge outside. (c) Yes, flux will change as the total charge inside the surface has changed. 40. (a) For points inside the balloon,E = 0, (b)E decreases as the surface densityof charge decreases . . 0 E ../ . , (c) E remains constant because 2 0 E . q / 4.. r , where r is the distance of the point fromthe centre of the balloon.

ELECTROSTATICS www.physicsashok.in 48 Reasioning Type Questions THE NEXT QUESTIONS REFER TO THE FOLLOWING INSTRUCTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R) is given just below it of the statements, mark the correct answer as � (A) If both A and R are true and R is the correct explanation of A. (B) If both A and R are true but R is not the correct explanation of A. (C) If A is true but R is false. (D) If both A and R are false. (E) If A is false but R is true. 1. Assertion (A) : If there exists colomb attraction between two bodies, both of them may not be charged. Reason (R) : Due to induction effects a charged body can attract a neutral body. 2. Assertion (A) : A small metal ball is suspended in a uniform electric field with an insulated thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the electric field. Reason (R) : X-rays emit photoelectrons and metal becomes negatively charged. 3. Assertion (A) : Electric current will not flow between two charged bodies when connected if their charges are same. Reason (R) : Current is rate of flow of charge. 4. Assertion (A) : The surface densities of two spherical conductors of radii r1 and r2 are equal. Then the electric field intensities near their surfaces are also equal. Reason (R) : Surface charged density = charge/area. 5. Assertion (A) : When charges are shared between two bodies, there occurs no loss sof charge, but there does occur a loss of energy. Reason (R) : In case of sharing of charges, the energy of conservation fails. 6. Assertion (A) : Two adjacent conductors, carrying the same positive charge have a potential difference between them. Reason (R) : The potential of a conductor depends upon the charge given to it. 7. Assertion (A) : Dielectric breakdown occurs under the influence of an intense light beam. Reason (R) : Electromagnetic radiations exert pressure. 8. Assertion (A) : The tyres of aircrafts are slightly conducting. Reason (R) : If a conductor is connected to ground, the extra charge induced on conductor will flow to ground. 9. Assertion (A) : Metallic shield in the form of a hollow shell, can be built to block an electric field. Reason (R) : In a hollow spherical shell, the electric field inside it is zero at every point. 10. Assertion (A) : A bird perches on a high power line and nothing happens to the bird. Reason (R) : The level of bird is very high from the ground. 11. Ametal sphere is suspended from a nylon thread. Initially, the metal sphere is uncharged . When a positively charged glass rod is brought close to the metal sphere, the sphere is drawn toward the rod. But if the sphere touches the rod, it suddenly flies away from the rod. Explain why the sphere is first attracted, then repelled. Level # 1. Objective Type Question 1. Two copper spheres of same radii one hollow and other solid are charged to th

e same potential then (A) both will hold same charge (B) solid will hold more charge (C) hollow will hold more charge (D) hollow cannot be charged 2. Through the exact centre of a hydrogenmolecule, an . -particle passes rapidly, moving on a line perpendicular to the internuclear axis. The distance between the two hydrogen nuclei is b The maximum force experienced by the . -particle is (A) 2 2 0 4 3 3.. e b (B) 2 2 0 8 3..e b (C) 2 2 0 8 3 3. . e b (D) 2 2 0 4 3..e b 3. A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x = +1 cm and C be the point on the y-axis at y = +1 cm. Then the potential at the points A, B and C satisfy (A) VA > VB (B) VA < VC (C) VA < VB (D) VA > VC

ELECTROSTATICS www.physicsashok.in 49 4. On an equilateral triangle of side 1 m there are three point charges placed at its corners of 1C, 2C and 3C. The work required to move these charges to the corners of a smaller equilateral triangle of side 0.5 m is (A) 19.8 x 1010 J (B) 39.6 x 1010 J (C) 9.9 x 1010 J (D) 4.45 x 1010 J 5. The potential field depends on x and y coordinates as V = (x2 � y2). Correspondence electric field lines in x.y plane as shown in figure. (A) (B) (C) (D) 6. Two circular rings A and B, each of radius a = 130 cm are placed coaxially with their axes horizontal in a uniform electric field E = 105 NC�1 directed vertically upwards as shown in Figure. Distance between centers of these rings A and B is h = 40 cm, ring A has a positive charge q1 .10.C while ring B has a negative charge of magnitude 2 q . 20.C . A particle of mass m = 100 gm and carrying a positive charge q . 10.C is released from rest at the centre of the ring A. Calculate its velocity when it has moved a distance of 40 cm. (A) v . 6 2ms.1 (B) v . 4 2ms.1 (C) v . 7ms.1 (D) v . 32ms.1 7. Three identical spheres each having a charge q and radius R, are kept in such a way that each touches the other two. the magnitude of the electric force on any sphere due to other two is (A) . . 2 0 1 3 4 4 q.. R . . .. .. (B) . . 2 0 1 2 4 4 q.. R . . .. .. (C) . . 2 0 1 2 4 4 R.. q . . . . . . (D) . . 2 0 1 3 2 4

q.. R . . .. .. 8. A system consists of a uniformly charged sphere of radius R and a surrounding medium filled by a charge with the volume density . .. r , where . is a positive constant and r is the distance from the centre of the sphere. Find the charge of the sphere for which the electric field intensity E outside is independent of r. (A) 0 . 2 . (B) 0 2 . . (C) 2.. R2 (D) None of the above 9. Three charges each of +q, are placed at the vertices of an equilateral triangle. The charge needed at the centre of the triangle for the charges to be in equilibrium is (A) 3.q (B) . 3q (C) 3q (D) . 3q 10. Hollow spherical conductor with a charge of 500.C is acted upon by a force 562.5 N. What is electric intensity at its surface? (A) zero (B) 1.125 x 106 N/C (C) 2.25 x 106 N/C (D) 4.5 x 106 N/C 11. A hemisphere of radius R is charged uniformly with surface density of charge . What will be the potential at centre? (A) 0 2 . R . (B) 0 4.. (C) 0 2.. (D) 0 43 . R .

ELECTROSTATICS www.physicsashok.in 50 12. A circular cavity is made in a conductor. A positive charge q is placed A B at the centre (A) The electric field at A and B are equal (B) The electric charge density at A = the electric charge density at B (C) Potential at A and B are equal (D) All the above. 13. An isolated metallic object is charged in vacuum to potential v0, its electrostatic energy being W0. It is then disconnected from the source of potential, its charge being left unchanged and is immersed in a large volume of dielectric, with dielectric constant k. The electrostatic energy will be. (A) kW0 (B) 0 Wk (C) 0 2Wk (D) W0. 14. A charge +q is fixed at each of the points x = x0, x = 3x0, x = 5x0 .... ad inf. on the x-axis, and a charge � q is fixed at each of the points x = 2x0, x = 4x0, x = 6x0 .... ad inf. Here x0 is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/( 4...r ). Then, the potential at the origin due to the above system of charges is. (A) Zero (B) 8 x log 2 q ..0 0 e (C) . (D) 0 0 e4 xq log 2 .. 15. Apositively charged thinmetal ring of radius R is fixed in the xy-plane with its centre at the originO. Anegatively charged particle P is released from rest at the point (0, 0, z0) where z0 > 0. Then the motion of P is. (A) Periodic, for all values of z0 satisfying 0 < z0 < . (B) Simple harmonic, for all values of z0 satisfying 0 < z0 . R (C) Such that P crosses O and continues to move along the negative z-axis towards z = � . (D) None of these 16. For an infinite line of charge having charge density . lying along x-axis, the work done in moving charge from C to A arc CA is. (A) log 2 2q e ..0 . (B) log 2 4q e ..0 .

(C) log 2 4q e ..0 . (D) 2log 1 2q e ..0 . 17. A particle A has charge +q and particle B has charge + 4q with each of them having the same mass m. When allowed to fall from rest through same electrical potential difference, the ratio of their speeds vA : vB will be (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 18. Three point charges q1, q2 and q3 are taken such that when q1 and q2 are placed close together to form a single point charge, the force on q3 at distance L from this combination is a repulsion of 2 units inmagnitude. when q2 and q3 are so combined the force on q1 at distance L is an attractive force of magnitude 4 units. Also q3 and q1 when combined exert an attractive force on q2 of magnitude 18 units at same distance L. The algebraic ratio of charges q1, q2 and q3 is. (A) 1 : 2 : 3 (B) 2 : � 3 : 4 (C) 4 : � 3 : 1 (D) 4 : � 3 : 2 19. An electric potential is given by V = k(xy), where k is a constant. A particle of charge q0 is first taken from (0, 0) to (0, a) to (a, a), then directly from (0, 0) to (a, a) and lastly from (0, 0) to (a, 0) to (a, a). If W1, W2 and W3 be the work done for the individual paths respectively then (A) W1 = W2 = W3 = � q0ka2 (B) W1 = W3 = � 2q0ka2 (C) W1 = W3 > W2 (D) W1 > W2 > W3

ELECTROSTATICS www.physicsashok.in 51 20. Four charges 2C, � 3C, � 4C and 5C respectively are placed at all the corners of a square. Which of the following statements is true for the point of intersection of the diagonals ? (A) Electric field is zero but electric potential is non-zero (B) Electric field is non-zero but electric potential is zero (C) Both electric field and electric potential are zero (D) Neither electric field nor electric potential is zero 21. Two metallic identical spheres A and B carrying equal positive charge + q are a certain distance apart. The force of repulsion between them is F. A third uncharged sphere of the same size is brought in contact with sphere. A and removed. It is then brought in contact with sphere B and removed. What is the new force of repulsion between A and B ? (A) F (B) 8 3F (C) 2F (D) 4F 22. An electron of mass me, initially at rest, moves through a certain distance in a uniform electric field in time t1. A proton of mass mp, also initially at rest, takes time t2 to move through an equal distance in this uniform electric field. Neglect the effect of gravity, the ratio t2 /t1 is nearly equal to (A) 1 (B) (mp /me)1/2 (C) (me /mp)1/2 (D) 1836 23. Eight dipoles of charges of magnitude are placed inside a cube. The total electric flux coming out of the cube will be (A) 08e . (B) 016 e . (C) 0 e . (D) Zero 24. A particle of mass m and charge q is released from rest in a uniform electric field E. The kinetic energy attained by the particle after moving a distance x is. (A) qEx2 (B) qEx2 (C) qEx (D) q2Ex 25. Two point charges +q and �q are held fixed at (�d, 0) and (d, 0) respectively of a x-y coordinate system. Then (A) The electric field E at all points on the x-axis has the same direction. (B) Work has to be done in bringing a test charge from . to the origin [1995] (C) Electric field at all points on y-axis is along x-axis. (D) The dipole moment is 2qd along the x-axis. 26. Three charges Q, +q and +q are placed at the vertices of a ring-angled isosceles triangle as shown. The net electrostatic a Q +q +q energy of the configuration is zero if Q is equal to [2000] (A) 1 2

.q

. (B) 2 2 2 . q . (C) �2q (D) +q 27. Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketches as in [2001] (A) (B) (C) (D) 28. Two equal point charges are fixed at x = -a and x = +a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to [2002] (A) x (B) x2 (D) x3 (D) 1 x

ELECTROSTATICS www.physicsashok.in 52 29. A metallic shell has a point charge �q� kept inside its cavity. Which one of the following diagrams correctly represents the electric lines of forces? [2003] (A) (B) (C) (D) 30. Six charges of equal magnitude, 3 positive and 3 negative are to be placed on PQRSTU corners of a regular hexagon, such that field at the centre is double that of what is would have been if only P Q R T S U O one +ve charge is placed at R. [2004] (A) +, +, +, �, �, � (B) �, +, +, +, �, � (C) �, +, +, �, +, � (D) +, �, +, �, +, � 31. A Gaussian surface in the figure is shown by dotted lime. q1 �q1 q2 The electric field on the surface will be [2004] (A) due to q1 and q2 only (B) due to q2 only (C) Zero (D) due to all 32. Three infinitely long charge sheets are placed as shown in figure. The electric field at point P is [2005] (A) 0 2 k� . . (B) 0 4 k� . . Z Z=a Z=�a Z=�2a P x �2. �.. (C) 0 2 k� . . . (D) 0 4 k� . . . 33. Two equal negative charges �q are fixed at points (0, �a) on y-axis. A positive charge Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will [1984] (A) execute simple harmonic motion about the origin. (B) move to the origin remain at rest

(C) move to infinity (D) execute oscillatory but not simple harmonic motion 34. A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to: [1987] (A) 2. Q (B) 4. Q (C) 4. Q (D) 2. Q 35. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of �3Q, the new potential difference between the same two surfaces is: [1989] (A) V (B) 2V (C) 4V (D) �2V 36. A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in Figure as [1996] (A) a (B) 2 (C) 3 (D) 4

ELECTROSTATICS www.physicsashok.in 53 37. A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field due to the sphere at a distance r from its center [1998] (A) increases as r increases, for r < R (B) decreases as r increases, for 0 < r < . (C) is discontinuous at r = R. (D) None of these 38. An ellipsoidal cavity is carved within a perfect conductor Figure. a positive charge q is placed at the centre of the cavity. The points A A B q and B are on the cavity surface as shown in the figure. Then [1999] (A) electric field near A in the cavity = electric field near B in the cavity (B) charge density at A = charge density at B (C) potential at A = potential at B (D) total electric field flux through the surface of the cavity is 0 q . . More Than One Choice Questions: 39. Three concentric spherical metallic shells A, B and C of radii a, b and c (a < b < c) have charge densities of . , .. , and . respectively. The potentials of A, B and C are: (a) . . 0 1 A V a b c . . . . . (b) 2 0 1 B V a b c c . . . . . . . . . . . (c) 2 2 0 1 C V a b c c c . . . . . . . . . . . (d) . . 0 1 A B C V V V a b c . . . . . . . 40. When positively charged spheer is brought near a metallic sphere, it is observed that a force of attraction exists between two. It means:

(a) metallic sphere is necessarily negatively charged. (b) metallic sphere may be electrically neutral. (c) metallic sphere may be negatively charged. (d) mothing can be said about charge of metallic sphere. 41. A conducting sphere of radius R has a charge. Then: (a) the charge is uniformly distributed over its surface, if there is no external electric field. (b) distribution of charge over its surface will be non-uniform, if an external electric field exists in the space. (c) electric field strength inside the sphere will be equal to zero only when no exxternal electric field exists. (d) potential at every point of the sphere must be same. 42. A small sphere of amss m and having charge q is suspended by a light thread, then: (a) tension in the thread may reduce to zero if anohter charged sphere is placed vertically below it. (b) tension in the thread may increase to twice of its original value if another charged sphere is placed vertically below it. (c) tension in the thread is greater than mg if another charged sphere is held in the same horizontal line in which first sphere stays in equilibrium. 43. Two point charge: Q and �Q/4 are separated by a distance x. Then: (a) potential is zero at a point on the axis which is x/3 on the right side of the charge �Q/4. (b) potential is zero at a point on the axis which is x/5 on the left side of the charge �Q/4. (c) electric field is zero at a point on the axis which is at a distance x on the right side of the charge �Q/4. (d) there exist two points on the axis, where electric field is zero. 44. Two equal and oppositely charged particles are kept some distance apart from each other. A spherical surface having radius equal to separation between the particles and concentric with their midpoint is considered. Then: (a) electric field is normal to the surface at two points only. (b) electric field is zero at no point. (c) electric potential is zero at every point of one circle only. (d) net electric flux through the surafce is zero.

ELECTROSTATICS www.physicsashok.in 54 45. Two identical charges +Q are kept fixed some distane apart. A small particle P with charge q is placed midway between them. If P is given a small displacement . , it will undergo simple harmonic motion if: (a) q is positive and . is along the line joining the charges. (b) q is positive and . is perpendicular to the line joining the charges. (c) q is negative and . is perpendicular to the line joining the charges. (d) q is positive and . is along the line joining the charges. 46. Select the correct statement(s): (a) Charge cannot exist without mass but mass can exist without charge. (b) Charge is conserved but mass is not. (c) Charge is independent of state of rest or motion. (d) None of these 47. Which of the following quantities do not depend on the choice of zero potential or zero potential energy? (a) Potential at a point. (b) Potential difference between two points. (c) Potential energy of a system of two charges. (d) Change in potential energy of system of two charges. Fill in the blanks: 1. Figure shows line of constant potential in a region in which an electric field is present. The value of the potential are written in brackets. Of the points A, B and C, the magnitude of the electric A B C (50 V) (40 V) (30 V) (20 V) (10 V) field is greatest at the point ............. [1984] 2. Two small balls having equal positive charges Q (coulomb0 on each are suspended by two insulating strings of equal length L (metre) from a hook fixed to a stand. The whole set up is taken in a satellite into space where there is no gravity (state of weightlessness). The angle between the two string is ............ and the tension in each string is ......... newtons. [1988] 3. A point charge q moves from point P to point S along the path PQRS (figure) in a uniform electric field E pointing parallel to the positive direction of the X-axis. The coordinates of the points X Y P Q R S E P, Q, R and S are (a, b, O), (2a, O, O) (a, �b, O) and (O, O, O) respectively. The work done by the field in the above process is given by the expression ..................... [1989] 4. The electric potential V at any point x, y, z (all in metres) in space is given by V = 4x2 volts. The electric field at the point (1m, 0.2 m) is ............... V/m. [1992] 5. Five point charges, each of value +q coul, are placed on five vertices of a regular hexagon of side L metres. the magnitude of the force on the point charge of �q coul. Placed at he centre of the hexagon 1

2 3 5 4 q q q q q �qL is .......... newton. [1992] 6 True / False : 6. The work done in carrying a point charge from one point to another in an electrostatic field depends on the path along which the point charge is carried. [1981] 7. Two identical metallic spheres of exactly equal masses are taken. One is given a positive charge Q coulombs and the other an equal negative charge. Their masses after charging are different. [1983] 8. A small metal ball is suspended in a uniform electric field with the help of an insulated thread. If high energy X-ray beam falls on the ball, the ball will be deflected in the direction of the field. [1983]

ELECTROSTATICS www.physicsashok.in 55 9. A ring of radius R carries a uniformly distributed charge +Q. A point charge �q is placed in the axis of the ring at a distance 2R from the center of the ring and released from rest. The particle executes a simple harmonic motion along the axis of the ring. [1988] Table Match 10. Match List I and List II and select the correct answer using the codes given below the lists: The electric fields due to various charge distribution are (q is the total charge on the body, . is the surface charge density, . is the linear charge density) List I List II I. At a distance x from the centre of a A. 2 2 0 1 2 E x x R . . . . . . . . . . . uniformly charged ring of radius R. The point is on the line passing through the centre of the ring and perpendicular to the plane of the ring II. At a distance x from the centre of a uniformly B. 0 2 E x . . . . charged ring of radius R. The point is on the line passing through the centre of the ring and perpendicular to the plane of the disc. III. At a distance x from an infinite sheet of C. 0 2 E . . . uniform distribution of charge IV. At a distance x from an infinite line of charge D. 2 2 3 2 0 1 4 E qx . R x . . . . . . . . (A) I-A, II-C, III-B, IV-D (B) I-D, II-A, III-C, IV-B (C) I-C, II-B, III-D, IV-A (D) I-B, II-D, III-A, IV-C Passage Type Questions THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE E is the electric field created by q1. V is the voltage at a given point in the field E. Assume that the

electric field created by q2 is negligible compared to E. k is Coulomb�s Law constant. m1 si the mass of q1, and m2 is the mass of q2. 1. E can best be described as � (A) constant (B) decreasing as r increases (C) increasing as r increases (D) increasing as q2 increases 2. What is the work done on q2 when it is moved at constant velocity along the distance d? (A) Zero (B) 1 Vq (C) 2 Vq (D) 2 Edq 3. Which of the following represents the work done on q2 when moved from its present position to a distance r from q1? (A) 1 2 12 kq q r (B) 1 2 kq q r (C) 1 2 2kq q r (D) 2 1 2 kq q r 4. If q1 and q2 have opposite charges, then when q2 is moved from its present location to a distance r from q1, the force on q2 due to q1 � (A) decreases by a factor of 4 (B) remains the same (C) increases by a factor of 2 (D) increases by a factor of 4 d r 2r q1 q2

ELECTROSTATICS www.physicsashok.in 56 5. If q1 is positive then when q2 is moved from its present location to a distance r from q1, the magnitude of the voltage experienced by q2 due to E� (A) decreases by a factor of 4 (B) remains the same (C) increases by a factor of 2 (D) increases by a factor of 4 6. The strength of the electric field E at r is � (A) half the field strength at 2r (B) the same as the field strength at 2r (C) twice the field strength at 2r (D) four times the field strength at 2r 7. If q1 and q2 are both positively charged and q2 is released, what is the maximum velocity that can be achieved by q2? (A) 1 2 2 kq q m r (B) 1 2 2 2 kq q m r (C) 1 2 2 2 2kq q m r (D) 1 2 2 2kq q m r 8. Electrostatic precipitators use electric forces to remove pollutant particles from smoke, in particular in the smokestacks of coal-burning power plants. One form of precipitator consists of a vertical hollowmetal cylinder with a thin wire, insulated from the cylinder, running along its axis. A large potential difference is established between the wire and the outer cylinder, with the wire at lower potential. This sets up a strong radial electric field directed inward, producing a region of ionized air near the wire. Smoke enters the precipitator at the bottom, ash and dust in the smoke pick up electrons, and the charged pollutants are accelerated toward the outer cylinder wall by the electric field. Suppose the radius of the central wire is 80.0 .m, the radius of the cylinder is 12.0 cm, and the potential difference between the wire and the cylinder is 60.0 kV. Assume that the wire and cylinder are both very long in comparison to the cylinder radius. (a) What is the electric-field magnitude midway between the wire and the cylinder wall? (b) What magnitude of charge must a 30.0- .g ash particle have if the electric field computed in part (a) is to exert a force ten times the particle�s weight? 9. A Geiger counter detects radiation such as alpha particles by using the fact that the radiation ionizes the air along its path. The device consists of a thin wire on the axis of a hollow metal cylinder and insulated from it.

A large potential difference is established between the wire and the outer cylinder, with the wire at higher potential; this sets up a strong radial electric field directed outward. When ionizing radiation enters the device, it ionizes a few air molecules. The free electrons produced ar accelerated by the electric field toward the wire and, on the way there, ionize many more air molecules. Thus a current pulse is produced that can be detected by appropriate electronic circuitry and converted to an audiable �click.� Suppose the radius of the central wire is 50.0.m and the radius of the hollow cylinder is 2.00 cm. What potential difference between the wire and the cylinder is required to produce an electric field of 6.00 x 104 N/C at a distance of 1.50 cm from the wire? (Assume that the wire and cylinder are both very long in comparison to the cylinder radius.

ELECTROSTATICS www.physicsashok.in 57 Level # 2 1. As shown a solid spherical region having a spherical cavity whose diameter R is equal to the radius of spherical region P C Q R that has a total charge Q. Find the potential at a point P, which is at a distance �x� from C. 2. A spherical water drop of radius a has a charge Q spread uniformly over its face. The drop is split into two identical spherical droplets (each of which has charge Q/2 spread uniformly over its face), which are kept very far from one another. (a) Compute the change in the electrostatic potential energy caused by the splitting. (b) Repeat the above calculation for the case in which the charge is uniformly distributed in the drop volume, before and after the splitting. 3. An electric dipole is placed at distance x from centre O on the axis of a charged ring of radius R and charge Q uniformly distributed over it. (a) Find the net force acting on the dipole. (b) What is the work done in rotating the dipole through 180°. (c) If the dipole is slightly rotated about its equilibrium position, find the time period of oscillation. Assume that the dipole is linearly restrained. 4. Particle 1 located far from particle 2 and possessing the kinetic energy T0 and mass m1 strikes particle 2 of mass m2 through the aiming parameter ., the arm of the momentum vector relative to particle 2 as; in the figure. Each particle carries a charge +q. 1 2 r P P0 . Find the smallest distance between the particles when m1 < < m2. 5. An electric field line emerges from a positive point charge +q1 at an angle . to the straight line connecting it to a negative point charge q2. At what angle . will the field + . . enter the charge � q q1 q2 2 ? 6. Two point charges each carrying a positive charge of 5e (e being the magnitude of the electronic charge) are separated by a distance 2d. An electron describes a circular path due to the attraction of the charges in a plane bise- A B d 5e 5e O d P R cting perpendicularly the line joining the two point charges. If the radius of the circular path described by the electron is R, determine the orbital speed of the electron. 7. Five thousand lines of forces enter a certain volume of space, and three thousand lines emerge from it.

What is the total charge in coulombs with in the volume ? 8. Determine the strength E of the electric field at the centre of a hemisphere produced by charges uniformly distributed with a density . over the surface of this hemisphere. 9. Two small identical balls lying on a horizontal plane are connected by a weight-less-spring. One ball (ball 2) is fixed and the other (ball 1 is free. The balls are charged identically as a result of which the spring length + � 12 increases . = 2 times. Determine the change in frequency. O 10. A semi-circular ring of mass M and radius R with linear charge density . hinged at its centre is placed in a uniform electric field as shown in the figure. (i) Find the net force acting on the ring. (ii) It the ring is slightly rotated about O and released, find its time period of oscillation (iii) What is the work done by an external agency to rotate it through an angle ..

ELECTROSTATICS www.physicsashok.in 58 11. Determine the force F of interaction between two hemispheres of radius R touching each other along the equator if one hemisphere is uniformly charged with a surface density, . 1 and the other with a surface density . 2. 12. Suppose in an insulating medium, having di-electric constant k = 1, volume density of positive charge varies with y-coordinate to law . = ay. A particle of mass m having positive charge q is placed in the medium at point A(0, y0) and projected with velocity v = v0 . i and assuming o x y v A 0 electric field strength to be zero at y = 0, calculate slope of trajectory of the particle as a function of y. 13. A metal ball of radius r and density . is charged by direct contact from the Earth�s surface till it acquires its maximum value. What should be this charge Qmax on the ball and the charge Q on the Earth (assuming it to be uniform sphere of mass m and radius R) such that it may be launched from the Earth�s surface with zero launch velocity. 14. A particle of mass m having negative charge q moves along an ellipse around a fixed positive charge Q so that its maximum and minimum distances from fixed charge are equal to r1 and r2 respectively. Calculate angular momentum L of this particle. 15. Three concentric, conducting spherical shells A, B and C have radii a = 10 cm, b = 20 cm and c = 30 cm respectively. The innermost shell A is earthed and charge q2 = 4 .C and q3 = 3 .C are given to shells B and C respectively. Calculate charge q1 induced on shell A and energy stored in the system. 16. The system consists of a hemispherical dielectric with volume charge density � . �. Find the potential difference between points A and B. . BA 17. A space is filled up with volume density of charge r 3 0 e . . . . . where .0 and . are positive constants, r is the distance from centre of the system. Find the magnitude of the electric field strength vector as a function of r. 18. Two identical balls are suspended from the same point by two threads. The balls are given equal charges and immersed in kerosene. Determine the density of the material of the balls if the threads deflect equally in vacuum and kerosene. The density of kerosene .0 = 0.8 g/cm3 and its relative permittivity .0 = 2. 19. A charge �q� is placed on the surface of an originally uncharged soap bubble of radius R0. Due to the mutual repulsion of the charged surface, the radius is increased to a somewhat large value R. Show that q = 1/ 2 20 0 2 0 0

2. PR R(R RR R ) 3 32 ... ... . . . . in which P is the atmospheric pressure. Level # 3 1. Two fixed charges �2Q and Q are located at the points with coordinates (�3a, 0) and (+3a, 0) respectively in the x-y plane. (a) Show that all points in the x-y plane where the electric potential due to the two charges is zero, lie on a circle. Find its radius and the location of its centre. (b) Given the expression V(x) at a general point on the x-axis and sketch the function V(x) on the whole xaxis. (c) If a particle of charge +q starts form rest at the centre of the circle, show by a short quantitative argument that the particle eventually crosses the circle. Find its speed when it does so. [IIT 91] 2. A blank to be filled appears in each of the following statements. Write in your answer book the subquestion number and write down against it your answer corresponding to each blank. In your answer, the sequence of the sub-questions should be the same as given in the question paper. (i) If .0 and .0 are respectively, the electric permittivity and magnetic permeability of free space, . and . the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is .....

ELECTROSTATICS www.physicsashok.in 59 (ii) The electric potential V at any point x, y, z (all in meters) in space is given by V = 4x2 volts. The electric field at the point (1m, 0, 2m) is ...... V/m. (iii) Five point charges, each of value +qC, are placed on five vertices of a regular hexagon of side L metre. The magnitude q q q q q of the force on the point charge of value � qC placed at the centre of the hexagon is ..... newton. [IIT 92] 3. A circular ring of radius R with uniform positive charge density . per unit length is located in the y-z plane with its centre at the origin O. A particle of mass m and positive charge q is projected from the point P(R 3, 0, 0) on the positive x-axis directly towards O, with an initial speed v. Find the smallest (non-zero) value of the speed v such that the particle does not return to P. 4. Consider the classical-model of an atom such that a nucleus of charge +e is uniformly distributed within a sphere of radius 2 Å. An electron of charge (� e) at a radial distance 1 Å moves inside the sphere. Find the force attracting the electron to the centre of the sphere. Calculate the frequency with which the electron would oscillate about the centre of the sphere. [REE 95 5. A charge 10�9 coulomb is located at origin in free space and another charge Q at (2, 0, 0). If the x-component of the electric field at (3, 1, 1) is zero calculate the value of Q. Is the y-component zero at (3, 1, 1) ? [REE 95] 6. A radioactive source in the form of ametal sphere of radius 10�2m, emits beta particles at the rate of 5 × 1010 particles per s. The source is electrically insulated. How long will it take for its potential to be raised by 2 volt assuming that 40% of the emitted beta particles escape the source ? [REE 97] 7. A non-conducting disc of radius a and uniform positive surface charge density . is placed on the ground, with its axis vertical. A particle of mass m and positive charge q is dropped, along the axis of the disc, from a height H with zero initial velocity. The particle has q/m = 4.0g/.. [IIT 99] (a) Find the value of H if the particle just reaches the disc. (b) Sketch the potential energy of the particle as a function of its height and find its equilibrium position. 8. A particle of charge q and mass m moves rectilinearly under the action of an electric field E = A � Bx where B is a + ve constant and x is a distance from the point where the particle was initially at rest. Calculate (a) Distance travelled by the particle till it comes to rest and (b) Acceleration at that moment. 9. Four point charges + 8 .C, � 1 .C, � 1 .C and + 8 .C are fixed at the points 27 / 2m, . 3 / 2 m, . 3/ 2m and . 27 / 2 m respectively on the y-axis. A particle of mass 6 × 10�4 kg and of charge + 0.1 .C moves along the � x direction. Its speed at x = + . is v0. Find the least value of v0 for which the particle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free. Given

1/(4..0 ) = 9 × 109 Nm2/C2. [IIT 2000] 10. Eight point charges are placed at the corners of a cube of edge a as shown in figure. Find the work done in disassembling this system of charges. [JEE 2003]

ELECTROSTATICS www.physicsashok.in 60 Answer Key Assertion-Reason Que. 1 2 3 4 5 6 7 8 9 10 Ans. A C D B C B B A A C Level # 1 Q. 1 2 3 4 5 6 7 8 9 10 Ans. A C A C A A A C A B Q. 11 12 13 14 15 16 17 18 19 20 Ans. A A B D A A B D A B Q. 21 22 23 24 25 26 27 28 29 30 Ans. B B D C C B C B C C Q. 31 32 33 34 35 36 37 38 39 40 Ans. D C D B A D A C BC BC Q. 41 42 43 44 45 46 47 Ans. ABD ABC ABC ABCD AC ABC BD Fill in the blanks, True-False & Match the column 1. B 2. 180°, 22 0 1 4 4Q .. L . 3. �qEa 4. �8 5. 2 0 1 4 q q .. L. 6. F 7. T 8. T 9. F 10. B Que. 1 2 3 4 5 6 7 Ans. B A A D C D D Passage Type 8. (a) 1.37 x 105 volt/m (b) 2.15 x 10�11 CLevel # 2 1. . . .. . . .. . . .. . 0 2 R2 4x2 1 x2 7 V Q 2. (a) The fractional change = � ... ..

. . . 22 / 3 1 1 (b) The fractional change in this case is also the same,. 3. (a) 2 2 2 2 5/2 0 aqQ R 2x 2 (R x ) . . . . .. . . . . (b) 2 2 3/ 2 0 aqQx .. (R . x ) 4. 2 2 0 02 min (1 1 (2 T / q ) 2Tr q . . . . . . , 4 0 1.. . .

ELECTROSTATICS www.physicsashok.in 61 5. sin / 2 qq 2 sin 21 . . ... ... . 6. 2 2 3 / 2 2 0 (R d ) V Re 5.. . . 7. 17.7 nc 8. 4.0 . 9. n 3n . 2 times 10. (i) ..RE (ii) 2 N 2 E . . (iii) 4.ER2 11. 1 2 02 2F R . . . . . 12. (y y ) 3m v a q 303 200 . . 13. 2 0 max Q 4 r Rg 3 .. . . , 3 0Q 4 r gR 3 .. . . 14. 2 (r r ) mr r Qq0 1 2 1 2 .. . 15. q1 = � 3 .C, Energy = 9.45 J 16. 0 3

02 3 V R 2 2 R .. . . . . 17. . r3 . 0 2 0 1 e 3 r. . .. .. 18. 1.6 gm/cm3 Level # 3 1. (a) radius = 4a, centre dt (5a, 0) (b) .. . .. . . . .. . . | x a | 2 | x 3a | 1 4V Q0 (c) At x = 9a where V = 0, the charged particle eventually crosses the circle, 8 ma V 9q ..0 . 2. (i) .0.0 .. (ii) � 8 V/m (iii) 2 9 2 L 9 .10 q (N) 3. 2 m V q.0 . . 4. 9 × 1014 Hz 5. Q = � 4.27 × 10�10 C, 0 4 11 11 E 2 10 0 9 y . ..

.

. .

. 6. 700 .F 7. (a) H = 34a and H = 0 (b) 3 a .0 . 8. x = 0, x = B2A , a = n . qA 9. Vmin = 3 m/s, K.E at origin = 2.5 × 10�4 J 10. W = 5.824 . .. . . .. . .. a . q 4 1 2 0 �X�X�X�X�

CURRENT ELECTRICITY

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

ELECTRIC CURRENT www.physicsashok.in 1 THEORY OF ELECTRICCURRENT CURRENT ELECTRICITY The time rate of flow of charge through any cross-section is called current. Therefore, if through a crosssection, .q charge passes in time .t, the average electric current through that area av i qt . . . and instantaneous current t 0 i lim q dq . . t dt . . . . . Regarding electric current following points areworth noting : (a) Current is assumed to be a fundamentalquantityin physicswith unit ampere and dimension [A].TheCGS unit of current is emu of current and is called biot (Bi), i.e., 1A = 1C = (1/10)emu of charge s s 1A = 1 Bi 10 (b) The current is same for allcross-sections of a conductor of non-uniformcross-section. Similar towater flow, charge flows fasterwhere the conductor is smaller in cross-section and slowerwhere the conductor is larger in cross-section, so that charge rate remains unchanged. (c) Though conventionallya direction is associatedwith current (opposite to the motion of electrons) it is not a vector as the directionmerely represents the sense of charge flow and not a true direction. Further current does not obey i1 i2 i = i1 + i2 the lawof parallelogramof vectors, i.e., iftwo currents i1 and i2 reach a point we always have i = i1 + i2 whatever be the angle between i1 and i2. (d) By convention, the direction of current is taken to be that inwhich positive chargemoves and opposite to the direction of flowof negative charge. (e) As charge is conserved and current is the rate of flowof charge, the charge entering at one end per second of a conductor is equal to the charge leaving the other end per second. (f) Current indifferent situations is due to motion of different charge carries. Current inconductors and vacuum tubes isdue tomotion ofelectrons, inelectrolytes due tomotion ofbothpositive and negative ions, indischarge tube due tomotion of positive ions and negative electrons and in semiconductors due tomotion of electrons and holes. Example 1. The current in awire varieswith time according to the relation i (3.0A) 2.0 A t s . . . . .. .. (a) How many coulombs of charge pass a cross-section of the wire in the time int

erval between t = 0 and t = 4.0 s? (b)What constant current would transport the same charge in the same time interval ? Sol. (a) i dq dt . . q 4 0 0 . dq . . idt . 4 0 q . . (3. 2t)dt. . 2 40 q . ..3t . t .. . 12 .16 . 28 C

ELECTRIC CURRENT www.physicsashok.in 2 (b) i q 28 7 A t 4 . . . Types of Electric Current According to itsmagnitude and direction current is usuallydivided into two types : (i) Direct current : If themagnitude and direction of current does not varywith time, it is said to be direct current (DC). Cell, battery or DC dynamo are its sources. (ii) Alternating current : If a current is periodic (with constant amplitude) and has half cycle positive and half negative, it is said to be alternating current (AC).ACdynamo is the source of it. NOTE : It is worthy to note here that rectifier converts AC into DC while inverter DC into AC. Rectifier Inverter DC AC Current Density Current densityJ . at at a point is defined as a vector havingmagnitude equalto current per unit area surrounding that point and normal to the directionof charge flowand directioninwhich current passes through that point. So, if at point P current .I passes normally through area .S as shown in Fig., current densityJ . at P will be given by S 0 J lim i n . . S . . . . . + � dS i P n J i.e., J di n dS . . . Regarding current densityfollowing points areworth noting : (a) If the cross-sectional area is not normal to the current, the cross-sectional area normal to the current in accordancewith Fig., will be dS . dS cos . . J di dScos . . i J dS i.e., di = JdS cos . dS cos

or di . J . dS . . i.e., i . . J . dS . . (b) In case of conductors asV= iRand by definitions, E V and R V L S . . . So, .EL. i LS . . i.e., J i 1 E S . . . or J . . E . .. . 1 . . . . . . . . .

ELECTRIC CURRENT www.physicsashok.in 3 (c) In case of uniformcharge flowthrough a cross-section normal to it as i = nqvS So, J i n (nqv) n S . . . . . or charge J . nq v . v (.) . . . [(.)charge = nq] Drift Velocity The average uniformvelocity acquired by free electrons inside ametalby the application of an electric field which is responsible for current through it is called drift velocity. It is represented by�vd�. The current flowing throughconductor d d i neAv or v i neA . . where n= number ofmoving electrons per unit volume A= area of cross-section NOTE : Some books have taken average drift velocity as half of initial and final drift velocities of an electron giving vd = (eE./2m). This is wrong as vd is the average of drift velocities of large number of electrons at same instant and as for each electron v .. d = a f with .. / . . . .. a= e E m constant , . . . . v v . . / . .. d a = e E m . Example 2. An n-type silicon sample ofwidth 4 × 10�3m, thickness 25 × 10�5 mand length 6 × 10�2 mcarries a current of 4.8mAwhen the voltage is applied across the length of the sample.What is the current density ? If the free electron density is 1022 m�3, then find howmuch time does it take for the electrons to travel the full length of the sample ?Given that charge on an electron e = 1.6 × 10�19C. Sol. Bydefinition, . . J i i S b d . . . [as s = (b × d)] So, . . . . 3 3 2 5 3 J 4.8 10 4.8 10 A/m 25 10 4 10

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. .

.

. . .

. . . and as in case of electric conduction inmetals J = nevd or d v J ne . So, 3 d 22 19 4.8 10 v 3 m/ s 10 1.6 10. . . . . . Hence, time taken by electron to travel the length L (= 6 × 10�2 m) of the conductor 2 d L 6 10 t 0.02 s. v 3 . . . . . Example 3. The area of cross-section, length and density of a piece of a metal of atomicweight 60 are 10�6m2, 1mand5 × 103 kg/m3 respectively. Find the number of free electrons per unit volume ifeveryatomcontributes one free electron.Also find the drift velocityof electrons in themetalwhen a current of 16Apasses through it. Given thatAvogadro�s number NA = 6 × 1023/mol and charge on an electron e = 1.6 × 10�19C.

ELECTRIC CURRENT www.physicsashok.in 4 Sol. As according toAvogadro�s hypothesis, A A A N m so n N N m N d as d m N M V VM M V . . . . . . . .. .. . . . . . 23 3 28 3 3 6 10 5 10 n 5 10 /m 60 10. . . . . . . Nowas each atomcontributes one electron, the number of electrons per unit volume ne = 1 × n = 1 × 5 × 1028 = 5 × 1028/m3 Further as here, 6 2 6 J i 16 16 10 A/m A 10. . . . . and as J = nevd . . . . 6 3 d 28 19 v J 16 10 2 10 m/ s ne 5 10 1.6 10 . . . . . . . . . . NOTE : � From this example it is clear that : An electron will take 1/(2 × 10�3), i.e., 500 s = (8.3 min.) to travel 1 m length of wire if it can. � If resistivity of metal . is taken to 2 × 10�8 ohm�m, the electric field inside the metal E = .J = (2 × 10�8) × (16 × 106) = 0.32 V/m and not zero as in electrostatics. RESISTANCE AND OHM�S LAW For somematerials, especiallymetals, at a given temperature, J . is nearlydirectly proportional to E.. and the ratio ofthemagnitudesE and J is constant.This ratio is called the resistivity(.) and this relationship is called the Ohm�s law.Thus, resistivity EJ . . SI units ofresistivity are .�m(ohm�metre).The reciprocalof resistivityis conductivity(.).

Thus, . . 1. Amaterial that obeysOhm�s lawreasonablywell is called an ohmic conductor or a linear conductor.Materials which show substantial departures fromOhm�slaw arecalled non-ohmic or non-linear. Resistance Suppose a conductingwire has a uniformcross-sectional areaAand length l as shown in Fig. Let Vbe the potentialdifference between the ends of the wire. If themagnitudes of the current densityJ . and the electric field E.. are JE A + � Vl i uniformthroughout the conductor, the total current i is given by i= JAand the potential differenceVbetween the ends isV= El. . V E i JA A . l . . l Here, A . l is constant for ohmicmaterials. This is called the resistance R.

ELECTRIC CURRENT www.physicsashok.in 5 Thus, R Vi . The resistanceR of a particular conductor is related to the resistivity . ofitsmaterial by R A . . l The equation, V = iR is often calledOhm�s law. NOTE : The equation R= Vi defines resistance R for any conductor, whether or not it obeys ohm�s law, but only when R is some constant we can correctly call this relationship Ohm�s law. Thus, for ohmic conductors V�i group is a straight line possing through origin. The slope of this line is equal to the resistance of the conductor. V i . R = V = tan i . Reciprocal of resistance is called conductance (G), i.e. G = 1 = i r V SI unit of G is ohm�1 which is called mho. Example 4. Two copper wires of the same length have got different diameters, whichwire has : (a) greater resistance and (b) greater specific resistance ? Sol. (a) For a givenwire,R A . . l i.e. R 1A . So, the thinner wirewill have greater resistance. (b) Specific resistance (.) is amaterial property. It does not depend on l orA. So, both thewireswillhave same specific resistance. Example 5. Awire has a resistance R.What will be its resistance if it is stretched to double its length ? Sol. Let Vbe the volume ofwire, then V =Al A . Vl Substituting this in R A . . l ,we have 2 R V . . l So, for given volume andmaterial, (i.e., Vand . are constant)

R . l2 When l is doubled, resistancewill become four times, or the newresistancewill be 4R.

ELECTRIC CURRENT www.physicsashok.in 6 VARIATION OF RESISTANCE WITH TEMPERATURE The resistance of a conductor varieswith temperature. The graph of variation of resistance of puremetalwith temperature is shown. Mathematically the dependence of resistance R on temperature is expressed as R0 0 T0 T R Slope = R R (T) = R O 0[1 + .(T � T0)] In this equationR(T) is the resistance at temperature T and R0 is the resistance at temperature T0, often taken to be 0ºC or 20ºC. The factor . is temperature coefficient of resistivity. Example 6. The resistance of thin silver wire is 1.0 ..at 20ºC. Thewire is placed in a liquid bath and its resistance rises to 1.2 ..What is the temperature of the bath ? ..for silver is 3.8 × 10�3 perº C. Sol. R (T) = R0[1 + .(T � T0)] Here, R (T) = 1.2 ., R0 = 1.0. ..= 3.8 × 10�3 perº C and T0 = 20ºC Substituting thevalues,we have 1.2 = 1.0[1 + 3.8 × 10�3(T � 20)] or 3.8 × 10�3 (T � 20) = 0.2 Solving this, we get T = 72.6º C Example 7. Aresistance Rof thermal coefficient of resistivity= . is connected in parallelwith a resistance = 3R, having thermal coefficient of resistivity= 2.. Find the value of .eff . Sol. The equivalent resistance at 0ºC is 10 20 0 10 20 R R R R R . . ...(i) The equivalent resistance at tºC is 1 2 1 2 R R R R R . . ...(ii) But R1 = R10 (1 + .t) ...(iii) R2 = R20(1 + 2.t) ...(iv) and R = R0(1 + .efft) ...(v) Putting the value of (i), (iii), (iv), (v) in eqn (ii), eff 54 . . . Example 8. (a) The current density across a cylindrical conductor of radius R varies according to the equation 0 J J 1 r R . . . . .. .. , where r is the distance fromthe axis. Thus the current density i

s a maximumJ0 at the axis r = 0 and decreases linearlyto zero at the surace r = R.Calculate the current interms of J0 and the conductor�s cross sectional area isA= .R2. (b) Suppose that instead the current densityis amaximumJ0 at the surace and decreases linearlyto zero at the axis so that 0 J J r . R . Calculate the current.

ELECTRIC CURRENT www.physicsashok.in 7 Sol. (a)We concider a hollowcylinder of radius r and thickness dr. The cross-sectional area of considered element is dA= 2.rdr The current in considerd element is 0 dI JdA J 1 r 2 rdr R . . . . . . .. .. or 0 dI 2 J 1 r rdr R . . . . . .. .. . R 0 0 I 2 J 1 r rdr R . . . . . . .. .. 2 0 0 R A I J J 3 3 . . . (b) R 2 0 0 r I 2 J dr R . . . 3 R 0 0 I 2 J r R 3 . . . . . . . . 3 0 0 2 J R 2A I J R 3 3 . . . Example 9.Anetwork of resistance is constructedwith R1&R2 as shown in the figure. The potential at the points 1, 2, 3,......, N are V1, V2, V3,........, VN respectivelyeach having a potential k time smaller than R2 R2 R2 R3 V0 R1 V VN N�1 R1 R1 V1= V0 k

I I1 V2= V0 k2 VN= V0 kN I´ I´1 I´2 I2 previous one. Find: (i) 12 RR and 23 RR in terms of k (ii) current that passes through the resistance R2 nearest to theV0 in termsV0, k&R3. Sol. R2 R2 R2 R3 V0 R1 V VN N�1 R1 R1 V1= V0 kI I1 V2= V0 k2 VN= V0 kN I´ I´1 I´2 (i) According to kcL, I = I1 + I2 or 0 0 0 0 0 2 1 2 1 V V V 0 V V k k k k R R R . . . . . or . . . . 0 0 0 2 1 2 1 k 1 V V k 1 V kR kR k R . . . .

ELECTRIC CURRENT www.physicsashok.in 8 . . .2 12 R k 1 R k. . Also, I´ = I´1 + I´2 or 0 0 0 0 N 2 N 1 N 1 N 1 1 2 1 3 V V V 0 V 0 k k k k R R R R . . . . . . . . . . After solving, 23 R k R k 1 . . (ii)Here, 0 1 1 2 2 V 0 V 0 k I R R. . . . . . 0 0 0 1 2 2 3 3 V V k 1 V I kR k k R k R k 1 . . . . . . .. . .. Example 10. Arod of length Land cross-sectionareaAlies along the x-axis between x= 0 and x= L.Thematerial obeysOhm�s lawand its resistivityvaries along the rod according to .(x) = .0 e�x/L. The end of the rod at x = 0 is at a potentialV0 and it is zero at x = L. (a) Find the total resistance of the rod and the current in thewire. (b) Find the electric potential in the rod as a function of x. Sol. (a) The resistance of considered element is 0 x / L dR dx e dx

A A . . . . . . 0 L x / L 0 x /L L0 0 R e dx L e A A . . . .. . . . . . . . x dx 0 1 0 L L 1 R e 1 1 A A e . . . . . . . .. . .. . . . . . . 0L e 1 R A e . . . . . . . . . . . . 0 0 0 0 I V 0 V AeV R R L e 1 . . . . . . (b) . E = .J or x /L 0 E . . e. J E = 0 x /L I Ie A A . . . , . . x / L 0 0 0 e AeV E A L e 1 . . . . . . . x /L 01 E e V L 1 e . . . . . 0 x E V V x. . . V0 � Vx = Ex

ELECTRIC CURRENT www.physicsashok.in 9 or . . . . x /L 1 0 x 0 1 V e e V V Ex 1 e . . .. . . . . ELECTRIC CELL An electric cell is a devicewhichmaintains a continuous flow of charge (or electric current) in a circuit by a chemicalreaction. In an electric cell, there are two rods of differentmetals called as electrodes. These electrodes are kept in a solution called electrolyte. On joining two electrodes by a wire the charge begins to flow in the wire, i.e., current flows inthewire. Inside the cell, a chemical reaction takes place in the electrolytewhichmaintains the charge onthe electrodes and the flowofcharge inthewire is continuouslymaintained. Thus, a cellconverts the chemicalenergyinto electricalenergy. Inthe light ofmodernviews inreference to a cell following terms need to be reviewed. Electromotive Force (EMF) The emf ofa cell is defined aswork done bythe cell inmoving unit positive charge in thewhole circuit including the cellonce. Therefore, ifWis thework done bya cell inmoving a charge q once around a circuit including the cell, emf E Wq . SI unit of emf is joule/coulomband is called volt. The emf of a cell in a circuit is taken to be positive, if circuit current inside a cell, is fromnegative to positive terminal, (i.e., cell is discharging) otherwise negative as shown inFig. E1 E2 i � + � + i E = E1 + E2 (a) E1 E2 i � + + � i E = E1 � E2 (b) E1 E2 i � + + � i E = E2 � E1 (c) NOTE : The term electromotive force is misleading introduced by Volta who thought it to be

force that causes the current to flow. Actually emf is not a force but work required to carry unit charge from lower potential to higher potential inside the cell. Internal Resistance (r) and Terminal Potential Difference The potential difference across a real source in a circuit is not equal to the emf of the cell. The reason is that chargemoving throughthe electrolyte ofthe cell encounters resistance.We cellthis the internalresistance ofthe source, denoted byr.As the currentmoves through r, it experiences an associated drops in potential equal to �ir�.Thus,whena current is drawnthrougha source, the potentialdifference between the terminalofthe source is, V = E � ir This can also be shown as below: E r i A B VA � E + ir =VB or VA � VB = E � ir Following three special cases are possible : (i) If the current flows in opposite direction (as in case of charging of a battery), thenV= E + ir (ii)V= E, if the current through the cellis zero

ELECTRIC CURRENT www.physicsashok.in 10 (iii) V= 0, if the cellis short circuited. This is because current in the circuit i Er . or E = ir Short Circuited E r . E � ir = 0 or V = 0 Thus,we can summarise it as follows : E r i V = E � ir or V < E E r i V = E + ir or V > E E r V = E if i = 0 E r i =ErV = 0 if short circuited COMBINATION OF RESISTANCES In Series Fig., represent a circuit consisting ofa source of emf and two resistors connected in series. AB R1 R2 V V1 V2 ABV R i Let equivalent resistor is R as shown. Then R = R1 + R2 This result can be readily extended to a network consisting of �n� resistors in series. . R = R1 + R2 + ....... + Rn In Parallel Fig., represents a circuit consisting of a source of emf and two resistors connected in paralle. AB V R1 ABV R ii1 i2 R2 i IfR be the equivalent resistance, then 1 2

1 1 1 R R R . . This result can be extended to a network consisting of n resistors in parallel. The result is

ELECTRIC CURRENT www.physicsashok.in 11 1 2 n 1 1 1 .......... 1 R R R R . . . . NOTE : In series combination, the same current flows through each resistance while in parallel combination, the voltage drop across each resistor is equal to the source voltage V. Example 11. Compute the equivalent resistance of the network shown in Fig., and find the current idrawn fromthe battery. 18V i i 4 6 3 Sol. The 6 . and 3 . resistances are in parallel. Their equivalent resistance is, 18V i 4 2 1 1 1 or R 2 R 6 3 . . . . Nowthis 2 . and 4 . resistances are in series and their equivalent resistance is 4 + 2 = 6 .. Therefore, equivalent resistance of the network = 6 .. i 18V 6 Current drawn fromthe batteryis, i = net emf = 18 = 3A net resistance 6 Star-Delta (.) Conversion For Fig. (a) . (Delta), to be equivalent to Fig. (b) (Star)Y, R2 A B C R1 R3 R1 R3 R2RA A B C RB RC A B C RA RB RC (a) (b) (c) 1 3 1 2 A B 1 2 3 1 2 3 R R R , R R R R R R R R R . . . . . .

ELECTRIC CURRENT www.physicsashok.in 12 and 2 3 C 1 2 3 R R R R R R . . . Example 12. Find the equivalent resistance betweenAand B in Fig. using Star-Delta Theorem. 10 A B CD 10 10 10 10 10 10 Sol. Using Star-Delta theorem, the equivalent circuit can be drawn as shown in fig. 10 10 10 RC = 5 RB = 5 RD = 2.5 15 22.5 5 9 5 A CD O B A O B A B CB CD C CB CD DB R R R 20 10 5 R R R 20 10 10 . . . . . . . . . . DC DB D DC DB CB R R R 10 10 2.5 R R R 10 10 20 . . . . . . . . . . BC BD B BC BD DC R R R 20 10 5 R R R 20 10 10 . . . . . . . . . . R´ 15 22.5 15 22.5 9 15 22.5 37.5

. .

. . . .

. Hence, RAB = R´ + 5 = 9 + 5 = 14 . Example 13.What willbe the change in the resistance of a circuit consisting of five identical conductors if two similar conductors are added as shown by the dashed line in figure. Sol.

ELECTRIC CURRENT www.physicsashok.in 13 r r r r r r r B A r r r r r r B A r 2r2r A r B A r r r B R2 = Req = 3r But before added the conductors, the equivalent resistance is 1 eq R . R´ . 5r . 21 R 3 R 5 . . KIRCHHOFF�S LAW Manyelectric circuits cannot be reduced to simple series-parallel combinations. For example, two circuits that cannot be so broken down are shown in fig. R1 R2 R3 E1 E2 A BD C F E (a) (b) E1 E2 E3 R1 R2 R3 R4 A B D E F G I H R5 C However, it is always possible to analyze such circuits byapplying two rules, devised byKirchhoff. First here are two terms that wewill use often. Junction : Ajunction ina circuit is a point where three ormore conductorsmeet. Junctions are also called nodes or branch points. For example, in figure (a) pointsDand Care junctions. Similarly, in figure (b) point B and F are junctions. Loop : Aloop is any closed conducting path. For example, in figure (a)ABCDA,DCEFDandABEFAare loops.

Similarly, in figure (b), CBFEC, BDGFBare loops. Kirchhoff�s rules consists of the following two statements. (i) Junction rule : The algebraic sumof the currents at any junction is zero. That is, junction . i . 0 i2 i2 i3 This lawcan also bewritten as, �the sumof all the curents directed towards a i4 point incircuit is equal to the sumof all the currents directed awayfromthat point.�

ELECTRIC CURRENT www.physicsashok.in 14 Thus, infigure i1 + i2 = i3 + i4 The junction rule is based on conservation of electric charge. (ii) Loop rule : The angebraic sumof the potentialdifferences in anyloop including those associated emf�s and those of resistive elements,must equal zero. That is, closed loop . .V . 0 The loop rule is based on the fact that the electrostatic field is conservative in nature. In applying the loop rule, we need sign convention as discussed below: (a) Whenwe travel through a source in the direction from� to +, the emf is considered to be positive. E A B Path (a) (b) Whenwe travel form+ to �, the emf is considered to be negative. E A B Path (b) (c) Whenwe travel through a resistor in the same direction as the assumed current, the iR termis negative because the current goes in the direction R A B Path (c) i of decreasing potential. (d) Whenwe travel through a resistor in the direction opposite to the assumed current, the iR termis positive because this represents a rise of potential. R A B Path (d) i Example 14. Find current in different branches ofthe electric circuit shown in figure. 2 A B CD 4 4 2 F E 2V 4V 6V Sol. ApplyingKirchhoff�s first law(junction law) at junctionB, i1 = i2 + i3 ...(i) ApplyingKirchhoff�s second lawin loop 1 (ABEFA), �4i1 + 4 � 2i1 + 2 = 0 ...(ii) ApplyingKirchhoff�s second lawin loop 2 (BCDEB), �2i3 � 6 � 4i3 � 4 = 0 ...(iii) Solving Eqs. (i), (ii) and (iii),we get i1 = 1A CD 2V 1 4V 2 6V

A BF E 2 4 4 2 i3 i2i1 i3 i1 2 i 8 A 3 . 3 i 5 A 3 . . Here, negative sign of i3 implies that current i3 is in opposite direction ofwhat we have assumed.

ELECTRIC CURRENT www.physicsashok.in 15 IMPORTANT FEATURES 1. Distribution of current in parallel connections : When more than one resistances are connected in parallel, the potential difference across themis equal and the current is i i2 3 i1 i R 2R distributed among themin inverse ratio oftheir resistance, as 3R i VR . or i 1R . for same value ofV e.g., inthe figure, 1 2 3 i : i : i 1 : 1 : 1 6 : 3 : 2 R 2R 3R . . 1 i 6 i 6 i 6 3 2 11 . . . . .. . . .. . 2 i 3 i 3 i 6 3 2 11 . . . . .. . . .. and 3 i 2 i 2 i 6 3 2 11 . . . . .. . . .. 2. Distribution of potential in series connections :Whenmore than one resistances are connected in series, the current through themis same and the potential distributed in the ratio of their resistance, as V = iR or V . R for same value of i. e.g., inthe figure, V1 V2 V3 R 2R 3R i V1 : V2 : V3 = R : 2R : 3R = 1 : 2 : 3 . 1 V 1 V V 1 2 3 6 . . . . .. . . .. 2 V 2 V V 1 2 3 3 . . . . .. . . .. and 3 V 3 V V 1 2 3 2 . . . . .. . . .. COMBINATION OF CELLS

Cells are usuallygrouped in following threeways : In Series Suppose �n� cells each of emfE and internal resistancer are connected in series as shown in figure. Then

ELECTRIC CURRENT www.physicsashok.in 16 r i R E E r E r Net emf = nE Total resistance = nr + R . Current in the circuit = Net emf Total resistance or i = nE nr +R NOTE : If palarity of m cells is reversed, then equivalent emf = (n � 2m)E while total resistance is still nr + R . .n - 2m. E i= nr+R In Parallel : Consider the following three cases : Ist case : Let �n�cells each of emfE and internal resistance �r�are connected in parallel. Net emf = E Total resistance r R n . . i R i E r E r E r . Current inthe circuit i = net emf total resistance or i = E R + r/n IInd case : Let �n�cells have different E and �r�. Net emf = Eeq . . . . E/r = 1/ r .. Total resistance = Req . . R 11/ r . . . i2 i3 i1 i R E1

E2 E3 r1 r2 r3 i A B F E Hence, C D eq eq E i R . or . . . . E / r i 1 R 1/ r . . . . IIIrd case : This ismost general case of parallel grouping inwhich E and �r�of different cells are different and the positive terminals of fewcells are connected to the negative terminals of the others as shown. Net emf = Eeq

ELECTRIC CURRENT www.physicsashok.in 17 . . . . . . 1 1 2 2 3 3 1 2 3 E / r E / r E / r 1 1 1 r r r . . . . . . . . . . . i i2 3 i1 i R E1 E2 E3 r1 r2 r3 i Total resistance = Req 1 2 3 R 1 1 1 1 r r r . . . . . . . . . . Hence, eq eq E i R . . . . . . . 1 1 2 2 3 3 1 2 3 E / r E / r E / r i 1 R 1 1 1 r r r . . . . . . . . . . . . In Mixed Grouping There are �n� identical cells in a rowand number of rows are �m�. Net emf = nE Total resistance = Req R nr m . . r i E R i

Hence, i nER nr m . . Example 15. Find the emf and internal resistance of a single batterywhich is equivalent to a combination of three batteries as shown in figure. 4V 10V 22 1 6V Sol. The givencombination consists of two batteries in parallelandresultant ofthese two inserieswiththe third one. For parallel combinationwe can apply, 1 2 1 2 eq 2 2 E E 10 4 E r r 2 2 3V 1 1 1 1 r r 2 2 . . . . . . . Further, eq 1 2 1 1 1 1 1 1 r r r 2 2 . . . . . . req = 1 . Nowthis is in serieswith the third one, i.e.,

ELECTRIC CURRENT www.physicsashok.in 18 1 1 6V 3V The equivalent emf of these two is (6 � 3)Vor 3Vand the internal resistancewill be, (1 + 1) or 2 .. r = 2 E=3V Example 16. Abatteryconsists of a variable number n of identical cells having internal resistance connected in series. The terminals of the batteryare short circuited and the current Imeasured. Which one of the graph belowshows the relationship between I and n ? (A) I/A O n (B) I/A O n (C) I/A O n (D) I/A O n (E) I/A O n Sol. nE nr I nE E nr r = = I n Example 17. In previous problem, if the cell had been connected in parallel (instead of in series) which of the above graphswould have shown the relationship between total current I and n ? (A) I/A O n (B) I/A O n (C) I/A O n (D) I/A O n (E) I/A O n Sol. En r I E nE r r n = = I nEr = I

n Example 18. Underwhat conditioncurrent passing through the resistanceR can be increased by short circuiting the battery of emfE2. The internal resistances of the two batteries are r1 and r2 respectively. E1 r1 E2 r2 R (A) E2r1 > E1(R + r2) (B) E1r2 > E2(R + r1) (C) E2r2 > E1(R + r2) (D) E1r1 > E2(R + r1) Sol. The current throughR before short circuit

ELECTRIC CURRENT www.physicsashok.in 19 1 2 1 2 I E E r r R = + + + After short circuit : 1 1I' E r R = + E2 E1 r1 r2 I� > I R 1 1 2 1 1 2 E E E r R r r R > + + + + E1r1 + E1r2 + E1R > E1r1 + E1R + E2r1 + E2R . 1 2 2 1 2 E r > E r + E R WHEATSTONE�S BRIDGE This is an arrangement of four resitances inwhich one resistance is unknown and rest known.TheWheatstone�s bridge is shown in fig. The bridge is said to be balancedwhen deflection in galvanometer is zero, i.e., ig = 0 and hence, the conditionof deflection is G R SP Q B A C E D i2 ig = 0 i1 i2 i1 i P R Q S . NOTE : In Wheatstone�s bridge, cell and galvanometer arms are interchangeable. G R S P Q B A C D G R S P Q B A C D

In both the cases, condition of balance bridge is, P = R Q S Example 19.Ahemisphere network of radius a ismade by using a conductingwire of resistance per unit length r. Find the equavalent resistance acrossOP. P B O D A C Sol. Point (AandC) and (DandB) are symmetricallylocatedwith respect to points Oand P. Hence, the circuit can be drawn as shown in figure. This is a balancedWheatstone bridge between P andO Hence, r3 can be removed.And, 1 2 PO R r r 4. . Here . . 1 PB PD a r r R R 2 . . . . r22 r22 r3 r21 r21 B C PO and r2 = ROB = (a)r

ELECTRIC CURRENT www.physicsashok.in 20 . . PO 2 ar R 8 . . . Ans. Example 20. In the circuit shown, what is the potential difference VPQ? (A) +3V (B) +2V 4V 2V Q P 1V 21 2 3 (C) �2V (D) none Sol. In ABCPA�2I + 4 � I + 2 = 0 I 6 2A 3 = = VP � VQ = �{algebriac sumof rise up and drop up of voltage} 2V 4V Q P 1V I A B C D II (I�I ) 1 I1 I1 I1 2 1 2 3 = � (2 � 2I) = � (2 � 4) . P Q V - V = 2 V Example 21. Two batteries one of the emf 3V, internal resistance 1 ohmand the other of emf 15 V, internal resistance 2 ohmare connected in series with a resistance R as shown. If the potential difference between a and b is zero a b R the resistance of R in ohmis (A) 5 (B) 7 (C) 3 (D) 1 Sol.According to loop rule, 3 � I � IR + 15 � 2I = 0 I 18 3 R = + . Va � Vb = � (�3 + I) a b

R 15V 3V I or 0 = 3 � I . I = 3 A . 3 18 3 R = + or 9 + 3R = 18 . R 9 3 3 . . . Example 22. Consider an infinite ladder network shown . in figureAvoltage V is applied between the points AandB. This applied velue of voltage is halved after each R1 R1 R1 R1 R1 R2 R2 R2 R2 R2 AB section. (A) R1/R2 = 1 (B) R1/R2 = 1/2 (C) R1/R2 = 2 (D) R1/R2 = 3 Sol. I = I1 + I2 1 2 1 v � v v � 0 v � v 2 2 2 4 R R R . . or 1 2 1 v v v 2R 2R 4R = +

ELECTRIC CURRENT www.physicsashok.in 21 or 1 2 1 1 1 v R R 2R = + I2 v v = 0 v/2 I I1 v/4 R2 R2 R2 R2 R2 v = 0 v = 0 v = 0 v = 0 v = 0 or 1 2 1 1 2R R = or 12 R 1 R 2 = Example 23. If the switches S1, S2 and S3 in the figure are arranged such that current through the batteryisminimum, find the voltage across pointsAand B. 6 3 6 S1 1 1 1 S3 S2 9 AB Sol. Forminimumcurrent through battery, equivalent resistance 24V across battery should bemaximum.Aswe know, in series resistance increases, but in parallel, resistance decreases. Fromthese points of view, all switches should be open. 9 6 6 3 1 A 7 A 24 V I I I0.5 B 4.5 1 1 24 V According to loop rule, 24 � 7I � 4.5I � 0.5I = 0 . I = 2A . VAB = VA � VB = 0.5 × I = 0.5 × 2 = 1 V Example 24. In the circuit shown in figure, calculate the following :

(i) Potentialdifference between points �a� and �b�when switch �S� is open. (ii) Current through�S� in the circuit when�S� is closed. 36 63 a S b 36V Sol. (i) Here 1 I 36 0 4 A 9. . . Also 2 I 36 0 4 A 9. . . Also, 36 � Va = I1 × 6 = 6 I1 or 36 � Va = 6 × 4 = 24 V . Va = 36 � 24 = 12 V 36 63 a b V = 0 V0 = 36V I1 I2 36 � Vb = 3 × 4 = 12 V and Vb = 36 � 12 = 24 V . Va � Vb = 12 � 24 = �12 V (ii) In loopABCIHA. �6(I � I1) + 3I1 = 0 or �6I + 6I1 + 3I1 = 0 or 9I1 = 6I

ELECTRIC CURRENT www.physicsashok.in 22 . 1 1 I 9 I 3 I 6 2 . . ...(i) In loop CDEJIC, 3 6 36 D CB A HI E J FI G (I � I1) I1 I2 (I � I1 + I2) �3(I � I (I1 � I2) 1 + I2) + 6(I1 � I2) = 0 or �3I + 3I1 � 3I2 + 6I1 � 6I2 = 0 or �3I + 9I1 � 9I2 = 0 ...(ii) In loop EFGHIJE, 36 � 3I1 � 6(I1 � I2) = 0 or 9I1 � 6I2 = 36 ...(iii) After solving eqn(i), (ii) and (iii) I2 = 3Afrom�b� to �a�. Example 25.An enquiring physics student connects a cell to a circuit andmeasures the current drawnfromthe cell to I1.When he joins a second identical cell is serieswith the first, the current becomes I2.When the cells are connected are in parallel, the current through the circuit is I3. Show that relation between the current is 3 I3 I2 = 2I1(I2 + I3) Sol. Let the equivalent resistance of circuit is R. The emf and internal resistance of cell is E and r respectively. . 1 I E r R . . and 2 I 2E 2r R . . and 3 I E r / 2 R . . L.H.S. = 3 2 . . 3I I 3E 2E r / 2 R 2r R . . . .. . .. . . .. . 2 3 2 6E 3I I

r / 2 R 2r R . . . R.H.S. = 2I1 (I2 + I3) 2E 2E E r R 2r R r / 2 R . . . . . . .. . .. .. . . .. . .. . 6E2 r / 2 R 2r R . . . Hence, L.H.S. = R.H.S. Example 26. Find the potential differenceVA �VB for the circuit shown in the figure. 1 1 1 1 1 1 1 1 1 1V 1V 1V 1V 1V 1V B 1V 1V A Sol. i1 + i2 = i 4i + i1 = 0

ELECTRIC CURRENT www.physicsashok.in 23 . x 49 . 4 B A 3 2 1 0 4i x ii2 i1 3i 2i x+4 x+3 x+2 x+1 i1 . A B V V 2 4 22 V 9 9 . . .. . . . . .. .. Example 27. Power generated across a uniformwire connected across a supply is H. If the wire is cut into n equal parts and all the parts are connected in parallel across the same supply, the totalpower generated in the wire is : (A) 2 Hn (B) n2H (C) nH (D) Hn Sol. Let the resistance of wire is R A = r . r = the resistance of each piece R nA n = r . = v2 H t R = ...(i) when all pieces are corrected in parallel, Then equivalent resistance is 0 2 r r R n n = = . 2 2 2 1 0 v n v H t t r R = = . n2H fromeqn. (i) Example 28. The ratio of powers dissipatted respectively in R and 3R, as shown is : (A) 9 (B) 27/4

3R 2R R (C) 4/9 (D) 4/27 Sol. P1 = Power dissiatted in resistance R 2 2I R 4 I2R 3 9 . . . . .. .. and P2 = Power dissipatted in 3R = I2(3R) = 3I2R 3R 2R R I/3 2I/3 I I . 2 1 2 2 P 4I R P 9 3I R . . . 12 P 4 P 27 = Example 29. In the figure shown the power generated in y is maximumwhen y = 5.. (A) 2 . (B) 6 . .......... 2R R y 10V, (C) 5 . (D) 3 . Sol. According to KOL : 10 � 2 I � I y � I R = 0 . I 10 2 y R = + + . P = I2y R y 10V I I I or 2 P 100 y (2 y R) = + + For maximumpower dissipatted,

ELECTRIC CURRENT www.physicsashok.in 24 dP 0 dy . . By solving, y = 2 + R or 5 = 2 + R . R . 3. Example 30. Find the current through 25Vcell&power supplied by20Vcell in the figure shown. 10 511 5 10V 5V 20V 30V 25V Sol. Fromfigure,1 I 55 0 5A 11. . . 2 I 5 0 1A 5. . . 10 511 5 10V 5V 20V 30V 25V 25V 25V 25V 25V 25V V=0 V=0 V=0 V=0 V=0 15V I4 I3 I2 I1 I=I +I +I +I4 1 2 3 4 30V 5V 55V 3 I 30 0 3A 10. . . 4 I 15 0 3A 5. . . . Electric current through 25Vcell = I1 + I2 + I3 + I4 = 5 + 1 + 3 + 3 = 12A The power supplied by 20Vcell is P = � 20I2= �20 × 1 = � 20 W Example 31. The current I through a rod of a certain metallic oxide is given by I = 0.2 V5/2, where V is the potential difference across it. The rod is connected in series with a resistance to a 6Vbattery of negligible internal resistance.What value should the series resistance have so that : (i) the current in the circuit is 0.44 (the value of (2.2)2/5 = 1.37)

(ii) the power dissipated in the rod is twice that dissipated in the resistance. Sol. (i) the potentialdifference across rod for current 0.44 is 20.44 5 V´ 0.2 . . . .. .. . .2V . 2.2 5 The potential difference across connected resistor isV´´= 0.44 R E = V´ + V´´ or . .26 . 2.2 5 . 0.44R or . .20.44R . 6 . 2.2 5 . . .26 0.2 5 R 0.44 0.44 . . R = 13.64 � 3.12 R = 10.52 . (ii)Total power supplied by battery is used byrod and resistor . E I = V I + I2 R

ELECTRIC CURRENT www.physicsashok.in 25 But VI = power dissipated in rod V I = 2I2R . EI = 2I2R + I2R = 3I2R or E = 3IR or 6 = 3IR . IR = 2 ...(i) Also, V´ + V´´ = E 2 I 5 IR 6 0.2 . . . . .. .. or 2I 5 2 6 0.2 . . . . .. .. or 2I 5 4 0.2 . . . .. .. or 2 5 5 I 4 2 32 0.2 . . . . . .. .. or I = 32 × 0.2 = 6.4A Fromeqn (i), IR = 2 . R 2 20 0.3125 6.4 64 . . . . Remarks : In the case of rod,V� I graph is not straight line. So, ohm�s lawis not applicable in the case of the givenrod. Example 32. Aperson decides to use his bath tubwater to generate electric power to run a 40watt bulb. The bath tub is located at a height of 10mfromthe ground&it holds 200 litres ofwater. Ifwe installawater driven wheelgenerator on the ground, at what rate should thewater drain fromthe bathtub to light bulb?Howlong canwe keep the bulb on, if the bath tubwas full initially. The efficiency of generator is 90%. (g = 10m/s�2) Sol. Power P gh dm dt . . or 40 gh dm 90 dt 100 . . or 40 0.9 gh dm dt . . dm 40 40 4 kg / sec.

dt 0.9 gh 9 10 9 . . . . . 0 m dm t dt . . . .. .. . 0 t m 200 450 sec. dm 4dt 9 . . . . . .. ..

ELECTRIC CURRENT www.physicsashok.in 26 Example 33. The circuit shown in figure ismade of a homogeneouswire of uniform cross-section. 1234 isasquare. Find the ratio Q12/Q34of the amounts of heat liberated per unit time in conductor 1�2 and 3�4. 3 4 1 2 Sol. Let us represent the central junction ofwires in the formof two junctions connected by thewire 5�6 as shown in figure. then in follows fromsymmetry that there is no current through thiswire. Therefore, the central junction can be removed fromthe initialcircuit. Further, R12 = R13 = R34 = R24 = r and 15 25 36 46 R R R R r2 . . . . Let Vbe the voltage between 1 and 2. 3 4 1 2 6 5 3 4 1 2 6 5 Then the amount of heat liberated in conductor 1 � 2. 2 12 V Q r . ...(i) Current through3 � 4, 3 4 . . i V r 2 3 . . . . . . 2 2 3 4 3 4 2 Q i r V r 2 3 . . . . . . . .2 1 2 3 4 Q 2 3 11 6 2 Q .. . . . . Ans. Metre Bridge Themetre bridge is the practicalapplication of theWheatstone�s network principle. G l (100 � l) D A C

P Q B R J S In such a bridge, the ratio of two resistances sayRand S, can be determined fromthe ratio of their balancing lengths. In Fig.,AC is a 1mlong uniformwire, Let AD = l cm, thenDC = (100 � l) cm Since, resistance . length . P AD Q DC 100 . . . l l If P is known, thenQcan be determined. Example 34. The potentiometer wire AB is 100 cm long. When AC = 40 cm, no deflection occurs in the galvanometer, findR.

ELECTRIC CURRENT www.physicsashok.in 27 G A B 10 R Sol. 10 AC R CB . . R 10 CB (10) 100 40 10 60 15 AC 40 40 . . . . . . . . . . . . . . . . . . . . . . . . Example 35. Abatteryof emf .0 = 10Vis connected across a 1 mlong uniform wire having resistance 10./m. Two cells of emf .1 = 2Vand .2 = 4Vhaving internal resistances 1 ..and 5 ..respectively are connected as shown in the figure. If a galvanometer shows no deflection at the point P, find the G 51 10 P A B 0 =10V 1=2V 2=4V distance ofpoint P fromthe pointA. Sol. The resistance inAP is R1 = x. = 10 x where xis inmetre. The resistance in PB is R2 = (1 � x)10 The equivalent circuit is According to loop rule 10 � IR1 � IR2 � 10I = 0 . 1 2 I 10 10 1 R R 10 10 10 2 . . . . . . G R1 P R2 I 10V I I A B E = r E1 r1 E2 r2 + = 56214+ 5 = 14 6 V r = 1 × 5 1 + 5 = 56But VA � VP = E = IR1 . 14 1 10x

6 2 . . . x 14 2 0.4667 m 46.67 cm 6 10 . . . . Example 36. In the figure shown for which values ofR1 and R2 the balance point for Jockey is at 40 cmfromA.WhenR2 is shunted by a resistance of 10 ., balance shifts to 50 cm. Find R1 and R2. (AB = 1m) : G R1 A B R2 Sol. Let resistance perunit lengthof potentiometer is ..Assume that P is contact point of potentiometer. . The resistance inAP is R3 = x. and the resistance in PB is R4 = (1 � x). ( . AB = 1 m) According to balance condition ofwheatstone bridge, R1R4 = R2R3 or R1(1 � x). = R2 x. or R1(1 � x) = R2x ...(i) whenR2 is shunted.

ELECTRIC CURRENT www.physicsashok.in 28 Then 2 2 2 2 2 R R 10 10R R 10 R 10 . . . . . . R1(1 � x). = R´2 x. . . 2 1 2 R 1 x 10R x R 10 . . . ...(ii) Fromeqn. (i) and (ii), 1 R 10 3 . . and R2 = 5.. IMPORTANT FEATURES 1. When batteryand galvanometer arms of aWheatstone�s bridge are interchanged, the balance position remains undisturbedwhile sensitivityof bridge changes. I � I G 1 R P I1 IG (I � I1 + IG) S I A B I (I1 � IG) Q 2. Two other common forms of balancedWheatstone�s bridge are shown. ABG P Q R S POTENTIOMETER Potentiometer is an ideal device to measure the potential difference betweentwo points. It consists of a long resistancewireAB of uniformcross-section inwhich a steadydirect current is set up bymeans of a battery. LE1 i2 = 0 A B i C G E2 , ri i Potentialgradient k Potential difference across AB Total length .

AB k VL . AB k iR i L . . .

ELECTRIC CURRENT www.physicsashok.in 29 where AB RL . . = resistance per unit lengthof potentiometerwire The emf of source balanced between pointsB and C CB 2 E . kl . i R . l l E2 = iRCB Applications (i) To find emf of an unknown battery : l E1 i2 = 0 A B i C G EK 1 E1 i2 = 0 A B i C G EU 2 1 l2 i We calibrate the device by replacing E2 by a source of known emfEK and then by unknown emfEU. Let the null points are obtained at lengths l1 and l2. Then, EK = i(. l1) and EU = i(. l2) Here, ..= resistance ofwireAB per unit length. . K 1 U 2 EE . ll or 1 U K 2 E E . . .. .. ll So, bymeasuring the lengths l1 and l2, we can find the emf of an unknown battery. (ii) To find the internal resistance of a call : Firstly the emfE of the cell is balanced against a lengthAD= l1. For this the switch S´ is left opened and S is closed.Aknown resistance R is then connected to the cell as shown.The terminalvoltageVis nowbalanced against a smaller lengthAD´ = l2.Here, no switch S Is opened and S´ is closed. Then,

12 EV . ll G (E, r) D´ D A S´ RE l2 S l1 Since, B E R r V R. . {. E = i(R + r) and V = iR} or 12 R r R. . ll or 12 r 1 R . . . . . . . . ll Example 37. Apotentiometer wire of length 100 cm has a resistance of 10 .. It is connected in series with a resistance and a cell of emf2Vand of negligible internal resistance.Asource of emf 10mVis balanced against a length of 40 cmof the potentiometer wire.What is the value of external resistance ?

ELECTRIC CURRENT www.physicsashok.in 30 Sol. Fromthe theoryof potentiometer, VCB = E, if no current is drawn fromthe battery or 1 CB AB E R E R R . . . . . . . . E1 E A B C G R i Here, E1 = 2V, RAB = 10 . CB R 40 10 4 100 . . . . . . .. .. and E = 10 × 10�3 V Substituting in above,we get R = 790 . MOVING COIL GALVANOMETER Moving coilgalvanometer is a device used to detect small current flowing in an electric circuit.With suitable modifications, it can be used to measure current and potentialdifference. Conversion of galvanometer into an Ammeter An ammeter is an instrument whichis used tomeasure current in a circuit in ampere (ormilli-ampere ormicroampere). Hence, it is always connected in series in the circuit. Since, the galvanometer coilhas some resistance of its own, therefore, to convert a galvanometer into an ammeter, its resistance is to be decreased so, to convert a galvanometer into ammeter a low resistance, called shunt (S) is connected is S a G b i Ammeter in parallelto the galvanometer as shown in figure. Here, g . . i S i S G . . and . .. . A . . S G R G S . . where RA = resistance of ammeter S = resistance of shunt G= resistance of galvanometer

Conversion of Galvanometer into Voltmeter AVoltmeter is an instrument which is used tomeasure the potential difference between two points of an electric circuit directly in volt (ormilli-volt ormicro-volt). Hence, it is connected in parallel across those two points of the circuit, betweenwhich the potentialdifference ig G R is to be measured.When it is connected. Voltmeter Since, the resistance of coilofgalvanometer of its ownis low, hence, to convert a galvanometer into a voltmeter, high resistance Ris series is connectedwith the galvanometer. Here, g i V R G . . where R +G= RV = resistance of voltmeter Example 38. Amoving coil galvanometer of resistance 10 . produces full scale deflection, when a current of 25

ELECTRIC CURRENT www.physicsashok.in 31 mAis passed through it. Describe showing full calculations, howwill you convert the galvanometer into a voltmeter reading upto 120V. Sol. Here, G = 10 .., ig = 25 mA= 25 × 10�3A To convert the galvanometer into voltmeter reading upto 120V: To convert a galvanometer into voltmeter of rangeV, a large resistance Rhas to be connected in series to it.The value ofR is given by g R V G i . . Here, V = 120 V . 3 R 120 10 4790 25 10. . . . . . Example 39. Amilliammeter of range 10 mAand resistance 9. is joined in a circuit as shown. Themetre gives full-scale deflection for current I whenAand B are used as its terminals, i.e., current enters at Aand leaves , 10 mA at B (C is left isolated). The value of I is A B C (A) 100 mA (B) 900 mA (C) 1 A (D) 1.1 A Sol. According to loop rule, �9 × 10 � 0.9 × 10 + 0.1 × (I � 10) = 0 or I 10 90 9 990 0.1 - = + = 10 mA I 10 mA I�10 I . I = 1000 mA . 1 A Example 40. In the circuit shown in figure reading of voltmeter is V1 when only S1 is closed, reading of voltmeter is V2 when only S2 is closed. The reading of voltmeter is V3 when both S1 and S2 are closed then V 6R 3R R E S2 S1 (A) V2 > V1 > V3 (B) V3 > V2 > V1 (C) V3 > V1 > V2 (D) V1 > V2 > V3 Sol. Step-I�When S1 is closed, I 4R . . . 1 v I 3R 3R 4R . . . . . R

3R 1 v 34 . . Step-II�When S2 is closed I 7R . . . 2 v . I . 6R R 6R 2 v 6R 6 7R 7 . . . . Step-III�When S1 and S2 both are closed. I 3R . . v3 = I × 2R R 2R

ELECTRIC CURRENT www.physicsashok.in 32 3 2 v 2R 3R 3 . . . . Example 41. In the circuit shown in figure the reading of ammeter is the samewith both switches openaswith both closed. then find the resistance R. (ammeter is ideal) + � 300 1.5V R 1000 50 A Sol. Step-I : Discuss the circuit when both switches open : According to loop rule : 1.5 � 300 I � 100 I � 50 I = 0 . I 1.5 15 1 A 450 4500 300 . . . Step-II :Discuss the circuit after closing the switch. In loopABCDEA �IR + 1.5 � 300I1 = 0 or 300I1 + IR = 1.5 ...(1) A 50 I I I I 100 300 1.5 V In loop BCGFB �100I + (I1 � I) R = 0 or (I1 � I)R = 100 I I1R = (100 + R)I . . . 1 100 R I I R. . ...(2) Fromeqn (1) and (2) .100 R. I 300 IR 1.5 R. . . 300 1.5V R 100 50 A no current I1 I1

I1 AFB I C DE (I1�I) G or .100 R. 1 R 300 1.5 R 300 300 . . . . R = 600 . Example 42. The battery in the diagramis to be charged by the generator G. The generator has a terminal voltage of 120 volts when the charging current is 10 amperes. The battery has an emf of 100 volts and an internal resistance of � + G R +� 1 ohm. In order to charge the battery at 10 amperes charging current, the resistance R should be set at (A) 0.1 . (B) 0.5 . (C) 1.0 . (D) 5.0 . Sol. VA � VB = � {algebraic sumof rise up and drop up of voltage} 120 = � {�IR � 1 × I � 100} 120 = IR + I + 100 100V I B A R I or 20 = 10R + 10 . R . 1 . Example 43. Agalvanometer having 50 divisions providedwitha variable shunt �s� is used tomeasure the current when connected inserieswith a resistance of 90 ..and a battery of internal resistance 10 .. It is observed that when the shunt resistance are 10., 50., respectivelythe deflection are respectively9&30 divisions.What is the resistance of the galvanometer ?

ELECTRIC CURRENT www.physicsashok.in 33 Sol. The electric current through galvanometer is proportionalto its deflection. I × . . gg I 9 3 I´ 30 10 . . But . . S g g S IR I R R . . Also, g S g S I ER R 100 R R . . . . . . S g g S g S g S ER I R R 100 R R R R . . . . . . . . . . or . . . . S g S g S g g Sg S g S g S g S R R R 100 R R I R R I´ R´ R R´ 100 R R´

R R´ . . . . . . . . . . . . . . . . . . . or . . . . . . g g g g g g 10 10R 100 R 10 3 R 10 10 50 R 50 50R 100 R 50 . . . . . . . . . . . . . . . . . . . . . . .. . .. . Rg = 233.3 .. IMPORTANT FEATURES 1. (a) The reading of an ammeter is always lesser than actual current in the circuit. E R i (a) E R i´ (b) A G i´ i´ S (c) For example, in Fig. (a), actual current throughRis, i ER . ...(i) while the current after connecting an ammeter of resistance A GS G S . . . .. . .. in serieswithRis, i´ E R A

.

. ...(ii) FromEqs. (i) and (ii), we see that i´< i and i´= i ,WhenA= 0, i.e., resistance of an ideal ammeter should be zero.

ELECTRIC CURRENT www.physicsashok.in 34 (b) Percentage error inmeasuring a current throughan ammeter is 1 1 i i´ 100 R R A 100 i 1R . . . . . . . . . . . . . . . . . . .. .. or % error A 100 R A . . . . .. . .. 2. (a) The reading of a voltmeter is always lesser than true value. VG R RV r i i i r For example, if a current �i�is passing through a resistance r, the actualvalue is, V = ir ...(i) Nowif a voltmeter of resistance RV(=G+R) is connected across the resistance �r�, the newvaluewill be . . VV i rR V´ r R . . . or V V´ ir 1 r R . . ...(ii) FromEqs. (i) and (ii), we can see that V´ < V and V´ = V if RV = . Thus, resistance of an idealvoltmeter should be infinite. (b) Percentage error inmeasuring the potentialdifference bya voltmeter is, V V V´ 100 1 100 V 1 r R . . . . . . . . . . . . . . . . . . . . . or V %error 1 100 1 r R . . . . . . . . . . . . . RC CIRCUITS

In precedingsectionswe dealt onlywith circuits inwhichthe current did not varywith time.Here,were discuss about time-varying currents. Charging of a Capacitor First we consider the charging of a capacitor without resistance. + �S V C + �V q = CV + � 0 Consider a capacitor connected to a battery of emfVthrough a switch S. On closing the switch, there is no time tag between connecting and charging.Acharge q0 = CVcomes in the capacitor as soon as switch is closed and q � t graph in this case is a straight line q0 q r parallel to t=axis. Ifwe employed a resistance in the circuit, charging takes some time Fig. The q � t equation in this case, q = q0(1 � e�t/.)

ELECTRIC CURRENT www.physicsashok.in 35 Here, q0 = CV SC C and . = CR= time constant V q � t graph is anexponentially increasing graph. The charge q increases expponentially from0 to q0. Fromthe graph and equationwe see that, At t = 0, q = 0 q0 q t 0.632q0 and at t = ., q = q0 At t = ., q = q0(1 � e�1) = 0.632 q0 Here, . can be defined as the time inwhich 63.2%charging is over in a C� R circuit. Current flows ina C� Rcircuit during charging of capacitor.Once charging as over or the steady state condition is reached the current becomes zero. The current at any time t can be calculated bay differentiating qwith respect to t. Hence, i0 i t Charging current is, i = i0e�t/. i.e., current decreases exponentiallywith time. Time i � t graph is as shown in fig. Discharging of a Capacitor Againwe consider the discharging of capacitor with resistance. Suppose a capacitor has a charge q0. The positive plate has a charge +q0 and negative plate �q0. S q = 0 + � q0 When the switch is closed, the extra electrons on negative plate immediately comes to the positive plate and net charge on both plates become zero. So, we cansay that discharging takes place immediately. S C R q0 + � In case ofC � Rcircuit discharging also takes time. The q � t equation in this case is, t / c 0 q . q e. . Thus, q decreases exponentially fromq0 to zero, as shown in Fig. At t = 0, q = q0 q0 q t 0.368q0 At t = ., q = 0 In case of discharging definition of . is charged. At time t = ., q = q0 e�1 = 0.368q0 Hence, in this case . can be defined as the timewhen charge reduces to 36.8%of itsmaximumvalue q0. During discharging current flows inthe circuit till�q� becomes zero.This current can be found bydifferentiating

�q�with respect to �t� butwithnegative signbecause charge is decreasingwith time.Hence, discharging current is, i = i0e�t/. This is an exponentially decreasing equation. Thus, i � t graph decreases exponentiallywith time fromi0 to 0. The i � t graph is as shown in Fig. i0 i t

ELECTRIC CURRENT www.physicsashok.in 36 Example 44. Acapacitor C= 100 µF is connected to three resistor each of resistance 1 k. and a battery of emf 9V.The switch S has been closed for long time so as to charge the capacitor.When switch S is opened, the capacitor discharges with 1k 1k 1k C9V S time constant (A) 33 ms (B) 5 ms (C) 3.3 ms (D) 50 ms Sol. 1k 1k at t = 0 C R = 0.5 k . Time constant . = RC = 0.5 × 103 × 100 × 10�6 = 50 × 10�3 S = 50 m/s Example 45. In the circuit shown in figure C1 = 2C2. Switch S is closed at time t = 0. Let i1 and i2 be the currents flowing throughC1 and C2 at any time �t�, then the ratio i1/i2 (A) is constant (B) increases with increase in time t C2 C1 V S RR (C) decreases with increase in time t (D) first increases then decreases Sol. Here 1 t RC 1 I V e R . . and 2 t RC 2 I V e R . . . 1 1 2 2 t RC t 1 1 1 R C C t

2 RC I e e I e . . . . . . . . . . . . . . 1 2 . 1 2 t C C 1 RC C 2 I e I . . . Hence option (B) is correct. Example 46. In the R�C circuit shown in the figure the total energy of 3.6 × 10�3 J is dissipated in the 10 . resistorwhen the switch S is closed. The initial charge on the 2µF 10 S capacitor is

ELECTRIC CURRENT www.physicsashok.in 37 (A) 60 µC (B) 120 µC (C) 60 2 µC (D) 602 µC Sol. According to conservationprinciple of energy: Total energy stored on capacitor appears as heat in resistor. . 20 3 q H 3.6 10 2C . . . . . or q02 = 2C × 3.6 × 10�3 q02 = 2 × 2 × 10�6 × 3.6 × 10�3 q02 = 14.4 × 10�9 . q0 = 12 × 10�5 q0 = 120 µC Example 47. The capacitors shown in figure has been charged to a potential difference ofVvolts, so that it carries a charge CVwith both the switches S1 and S2 remaining open. Switch S1 is closed at t = 0.At t = R1C switch S1 E S2 C + � R1 R2 S1 is opened and S2 is closed. Find the charge on the capacitor at t = 2 R1C + R2C. Sol. When t < R1C 1t q CV e R C . . At t0 = R1C, 1 0 q CV e CV e . . . when S2 is closed, q0 �q0 R1 R2 t = t E 0 At instant t (t > t0) . . 1 2 E q I R R 0 C . . . . or (R1 + R2)CI = EC � q or . . 1 2 R R C dq EC q dt . . . �q R1 R2

E + +q I I or . . 0 1 q t 1 2 CV t R C e R R C dq dt EC q . . . . . . or . . . . . . 1 q t 1 2 CV R C e R . R C...ln EC. q .. . t or . . . . 1 2 1 CV R R C ln EC ln EC q t R C e . . . . . . . . . . . . . . . . . . or . . 1 2 1 EC CV R R C ln e t R C EC q . . . . . . . . . . . . . . . But t = 2R1C + R2C Putting the value,

ELECTRIC CURRENT www.physicsashok.in 38 2 q CE 1 1 CV e e . . . . . .. .. Example 48. Inthe figure shown initiallyswitch is open for a long time. Nowthe switch is closed at t = 0. Find the charge on the rightmost capacitor as a function of time given that it was initiallyunchanged. V R R C C S Sol. Step-I :Discuss the circuit,when switch is open. Here I V 2R . Also, 0 q CIR CV 2 . . II I +q �q0 Step-II : Discuss the circuit after closing the switch : 1 V I R q 0 2 2 C . . . Also, . . 0 1 V R q q q I 0 2 2 C . . . . . or 0 1 V I R q q q 0 2 2 C C C . . . . . V R R V2R2 q q �q1 q1 q +q�q0 1 �(q +q�q0 1) or 0 V I R q q V IR 0 2 2 C C 2 2 . . . . . . or 0 V IR q q 0 C C . . . . or . . 0 q q IR V C.

. . or 0 dq q q R V dt C. . . . . . . . . or q t 0 0 0 Rdq dt V q q C . . . . . . . . . . . or q 0 0 q q RC ln V t C . . . . . .. . . . . . . .. . . . . . .. or 0 0 V q q RC ln C t V qC . . . . . . . . . . . . . . . . . . . .. .. or 0 t RC 0 V q q C e V qC . . . . .

ELECTRIC CURRENT www.physicsashok.in 39 or t V q0 q V q0 eRC C C . . . . . . . . . . . . . . . . . ...(i) But potentialdiffer accross each capacitors are sume. 0 1 1 q q q q C C . . . or q0 + q = 2q1 . 0 1 q q q 2. . Putting the value of q, t RC 1 q CV 1 1 e 2 2 . . . . . . . . . THINKING PROBLEMS 1. Is a current-carrying conductor electrically charged? 2. Is there an electric field near the surface of a conductor carrying direct current? 3. Is current a scallar or vector? 4. A potential difference V is applied to a copper wire of diameter d and length l.What is the effect on the electron drift velocity of (a) doubling l, (b) doubling V, and (c) doubling d? 5. A current i enters the top of a copper sphere of radius R and leaves through the diametrically opposite point.Are all parts equally effective in dissipating Joule heat? 6. Account for the increase in the resistance ofmetals with rise in temperature. 7. Answer briefly� how can three resistances of values 2., 3. and 6. be connected to produce an effective resistance of 4. ? 8. How can an electric heater designed for 220V be adopted for 110V without changing the length of the coil and also without a change in the consumed power ? 9. The brilliance of lamps in a roomnoticeably drops as soon as a highpower electric iron is switched on and after a short interval, the bulbs regain their original brilliance. Explain. 10. Acurrent is passed through a steelwire which gets heated to a dull red. Then half the wire is immersed in cold water. The portion out of the water becomes brighter.Why? 11. For manual control of the current of a circuit, two rheostats in parallel are preferable to a single rheostat. Why? 12. In a hollow nonconducting pipe, there are two streams of ions in opposite directions. The ions of one streamare negatively charged and constitute a current of strength l and those of

the other stream are positively charged and constitute a current of the same strength.What is the total current through the pipe? 13. The drift velocity of electrons is quite small. How then does a bulb light up as soon as the switch is turned on, although the bulbmay be quite far fromthe switch? 14. Consider a voltaic cell in the open circuit with the copper plate at a higher potential with respect to the zinc plate. The electrolyte between the plates is a conducting medium. So the charges on the plates should be neutralized at once as it happens when two charged conductors are joined by a wire.Why are the charges not neutralized immediately? 15. Does emf have electrostatic origin? 16. Why is it easier to start a car engine on a warmday than on a chilly day? 17. The resistance of the human body is about 10 k.. If the resistance of our body is so large, why does one

ELECTRIC CURRENT www.physicsashok.in 40 experience a strong shock from a live wire of 220 V supply? 18. Lay people have the notion that a person touching a high power line gets �stuck� to the line. Is this true? If not, what is the fact? 19. When a direct current flows through a conductor, the amount of energy liberated is VQ, where Q is the charge passing through the conductor and V is the potential difference, while an energy of VQ/2 is liberated when a capacitor is discharged.Why? 20. An ordinary cellwith a small emf can produce larger current than an electrostaticmachine which generates thousands of volts? Explain. THINKING PROBLEMS SOLUTIONS 1. No, the amount of positive and negative charge in any elementary volume remains the same through the electrons are inmotion. So a current-carrying conductor is electrically neutral. 2. There is an electric field inside the conductor. This is equal to the rate of fall of potentialalong the conductor. There is continuity of this field outside the surface of the conductor. 3. Current is a scalar but current density is a vector. 4. u = j/ne = .E/ne where s is the conductivity and E is the electric field (byOhm�s law j = .E), or u = (./ne) (V/l) (. E = V/l). Obviously from this expression (a) u is halved if l is doubled, (b) u is doubled if V is doubled, and (c) u remains uncharged on doubling d. 5. No, the resistance of elements at right angles to the diameter varies fromelement to element but current is the same through all sections and so heating effect varies fromsection to section. 6. metals have mobile electrons.With increase of temperature, the lattice vibrations increase in amplitude. Thus, the probability of electrons striking ionic cores increases. This amounts to increase in resistance with temperature. 7. Connect the 3. and 6. resistors in parallel and the 2. resistor in series with the combination of 3. and 6.. Then 3. and 6. in parallel will sumup to 3 6 2 3 6 . . . . and this, with 2. in series, willwork out to 4.. 8. Join the ends of the coil and apply the supply voltage (110V) between this common terminal and the midpoint of the coil as shown in the figure. Let P be the power of the entire coil when the coltage applied is V volts. TheP = V2/R where R is the resistance of the coil. When it is connected as shown in the figure, the two parts are in parallel and the resistance of each part becomes R/2. Therefore, P1 = (V/2)2/(R/2)=P/2 Total power = P/2 + P/2 = P 9. The cold resistance of the coil of the iron ismuch smaller thanwhen it is hot. So when the iron is switched

on, there is a large drop of voltage and consequently the bulbs do not receive the proper voltage.As the coil gets heated, its resistance increases and so the voltage drop is made up when it is fully heated. This is why after some time the bulbs region their brilliance. 10. The resistance of the immersed portion decreases and so the current through the entire wire increases. This is why the portion outside the water becomes brighter. 11. Suppose, to produce a certain current i in a circuit, the length of the single rheostat wire required is l. Then i = E/lr where E is the emf of the cell and r is the resistance per unit length of the wire. To produce the same current in the same circuit by using two rheostats in parallel, let l� be the length of each rheostat wire. Then i = E/(l�r/2) = 2E/l�r. Obviously, l�= 2l. Thusmore length of eachwire is required to produce the same current. This is definitely of advantage. 12. The current constituted by the negative ionsmoving opposite to the positive ions has the same sense as the current constituted by the positive ions. Hence the total current is 2I and not zero. 13. For the bulb to light up, it is not necessary for the same electron to travel from the switch to the bulb. When the switch is turned on, every electron in the circuit begins to move simultaneoulsy, including those in the filament of the bulb. 14. The charges are prevented frombeing neutralized because the electrostatic field due to the charges on the plates is opposed by the charges of ions that migrate into the solution.

ELECTRIC CURRENT www.physicsashok.in 41 REASONINGTYPE QUESTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R) is given just below it of the statements, mark the correct answer as � (A) If both A and R are true and R is the correct explanation of A. (B) If both A and R are true but R is not the correct explanation of A. (C) If A is true but R is false. (D) If both A and R are false. (E) If A is false but R is true. 1. Assertion (A) : When a wire is stretched so that its thickness is halved, its resistance would become 16 times. Reason (R) : The data is insufficient to predict. 2. Assertion (A) : A current flows in a conductor only when there is an electric field within the conductor. Reason (R) : The drift velocity of electron in presence of electric field decreases. 3. Assertion (A) : A current carrying wire should be changed. Reason (R) : The current in a wire is due to flow of free electrons in a definite direction. 4. Assertion (A) : The wires supplying current to an electric heater are not heated appreciably. Reason (R) : Resistance of connected wires is very small and H . R . 5. Assertion (A) : In meter bridge experiment, a high resistance is connected in series with the galvanometer. Reason (R) : As resistance increases current through the circuit increases. 6. Assertion (A) : A 60 watt bulb has greater resistance than a 100 watt bulb. Reason (R) : V 2 P R . 7. Assertion (A) : The conductivity of an electrolyte is very low as compared to ametal at room temperature. Reason (R) : The number density of free ions in electrolyte is much smaller compared to number density of free electrons in metals. Further, ions drift much more slowly, being heavier. 8. Assertion (A) : Terminal voltage of a cell is greater than emf of cell, during charging of the cell. Reason (R) : The emf of a cell is always greater than its terminal voltage. 9. Assertion (A) : Material used in the construction of a standard resistance is constantan or manganin. Reason (R) : Temperature coefficient of these materials is very small. 10. Assertion (A) : If the current of a lamp decreases by 20%, the percentage decrease in the illumination of the lamp is 40%. Reason (R) : Illumination of the lamp is directly proportional tot he current through lamp. 11. Assertion (A) : Heater wire must have high resistance and high melting point. Reason (R) : If resistance is high, the electrical conductivity will be less. 12. Assertion (A) : The range of given voltmeter can be increased. Reason (R) : By adjusting the value of resistance in series with galvanometer the range of voltameter can be adjusted.

ELECTRIC CURRENT www.physicsashok.in 42 LEVEL # 1 1. The resistance of tungsten increases with increasing temperature. As a result, the relation between the current, . , flowing in the tungsten filament of an electric lamp and the potential difference, V, between its ends is of the form. (A) IO V (B) IO V (C) I O V (D) IO V 2. The current in a copper wire is increased by increasing the potential difference between its ends. Which one of the following statements regarding n, the number of charge carriers per unit volume in the wire, and Vd, the drift velocity of the charge carriers, is correct? (A) n is unaltered but vd is decreased. (B) n is unaltered but vd is increased. (C) n is increased but vd is decreased. (D) n is increased but vd is unaltered. 3. A potentiometer is to be calibrated with a standard cell using the circuit shown in the diagram. The balance point is found to be near L. To improve accuracy the balance point should be nearer M. This may be achieved by (A) replacing the galvanometer with one of lower resistance. (B) replacing the potentiometer wire one of higher resistance per unit length. (C) putting a shunt resistance in parallel with the galvanometer. (D) increasing the resistance R. 4. When the switch 1 is closed, the current through the 8. resistance is 0.75 A. When the switch 2 is closed (only), the current through the 2. resistance is 1 A. The value of . is (A) 5 V (B) 5 2 V (C) 10 V (D) 15 V 5. The time constant . for the shown RC circuit is (A) RC (B) 2 RC (C) 2 RC (D) not defined R R . 6. Two cells of equal emf of 10 V but different internal resistances 3. and 2. are connected in series to an external resistance R. The value of R that makes the potential difference zero across the terminals of one of any cells is (A) 5. (B) 6. (C) 1. (D) 1.5. 7. The resistor in which the maximum heat is produced is given by (A) 2. (B) 3. (C) 4. (D) 12.

8. n resistances each of resistance R are joined with capacitors of capacity C (each) and a battery of emf E as shown in the figure. In steady state condition ratio of charge stored in the first and last capacitor is E R C R C R (A) n : 1 (B) (n�1) : R (C) (n2 + 1) : (n2 � 1) (D) 1 : 1

ELECTRIC CURRENT www.physicsashok.in 43 9. What is the equivalent capacitance between A and B in the circuit shown. (A) 6.F (B) 1.5 .F (C) Zero (D) 2.F 10. A resistance R = 12. is connected across a source of emf as shown in the figure. Its emf changes with time as shown in the graph. What is the heat developed in the resistance in the first four second. R = 12 . . Source . (Volt) 24 4 t(s) (A) 72 J (B) 64 J (C) 108 J (D) 100 J 11. A source of constant potential difference is connected across a A B conductor having irregular cross-section as shown in the Figure. Then P Q (A) electric field intensity at P is greater than that at Q (B) rate of electric crossing per unit area of cross section at P is less than that at Q (C) the rate of generation of heat per unit length at P is greater than that at Q (D) mean kinetic energy if free electorn at P is greater than that at Q 12. Suppose a voltmeter reads the voltage of a very old cell to be 1.40 V while a potentiometer reads its voltage to be 1.55 V. What is the internal resistance of the cell (A) 20. (B) 30. (C) 10. (D) 40. 13. In the above question, What is the current it would supply to a 5. resistor. Assume the voltmeter resistance be be 280.. (A) 44 A (B) 0.044 A (C) 4.4 A (D) None of the above. 14. If 1 . , 2 . and 3 . are conductances of three conductors then their equivalent conductance when they are joined in series will be (A) 1 . + 2 . + 3 . (B) 1 2 3 1 1 1 . . . . . (C) 1 2 3 1 2 3 . . . . .. .. (D) None of these 15. A conductor is made of an isotropic material (resistivity .) has rectangular cross-section. Horizontally dimension of the rectangle decreases linearly from 2x at one end to x at the other end and vertial dimension increases from y to 2y as shown in Figure. Length of the conductor along the axis is equal to . . A battery is connected across this conductor then (A) resistance of the conductor is equal to 4.. 9xy . (B) rate of generation of heat per unit length is maximum at middle cross-section. (C) drift velocity of conduction electrons is minimum at middle section. (D) at the ends of the conductor, electric field intensity is same.

16. Find the current through 4. resistor just after making the circuit (A) 0 A (B) 6 A (C) 12 A (D) 2 A 17. 1 m long metallic wire is broken into two unequal parts P and Q. P part of the wire is uniformly extended into another wire R. Length of R is twice the length of P and the resistance R is equal to that of Q. Find the ratio of the resistance of P and R (A) 1 : 4 (B) 1 : 3 (C) 1 : 2 (D) 1 : 1

ELECTRIC CURRENT www.physicsashok.in 44 18. Which of the following statements is/are correct for potentiometer circuit (A) Sensitivity varies inversely with length of the potentiometer wire (B) Sensitivity is directly proportional to potential difference applied across the potentiometer wire. (C) Accuracy of a potentiometer can be increased only by increasing length of the wire (D) Range depends upon the potential difference applied across the potentiometer wire. 19. In the given circuit the ammeter reading is zero. What is the value of resistance R ? (A) R = 100. (B) R = 10. (C) R = 0.1. (D) None of these 20. What is the equivalent resistance between A and B in the given circuit diagram. (A) 2. (B) 12. (C) 20. (D) 10. 21. A cell of emf. 1.5 V and internal resistance 0.5. is connected to a (non-linear) conductor whose . -V graph is shown in Figure. Find the current drawn from the cell and its terminal voltage (A) 1.5 A and 2 V (B) 1 A and 1 V (C) 1 A and 2 V (D) 2 A and 1.5 V 22. You are given several identical resistances each of value R = 10. and each capable of carrying a maximum current of 1A. It is required to make a suitable combination of these resistances to produce a resistance of 5. which can carry a current of 4 A. The minimum number of resistances required is (A) 4 (B) 10 (C) 8 (D) 20 23. A conductor or area of cross section A having charge carriers, each having a charge q is subjected to a potential V. the number density of charge carriers in the conductor is n and the charge carriers (along with their random motion) are moving with a velocity v. If . is the conductivity of the conductor and . is the average relaxation time, then (A) . . . nq2 m (B) nq2 m. . . (C) 2 nq2m. . . (D) 2 2nq m. . . 24. A vacuum diode consists of plane parallel electrodes separated by a distance d and each having an area A. On applying a potential V to the anode with respect to the cathode a current I flows through the diode. Assume that the electrons are emitted with zero velocity and they do not change the field between the electrodes. The electron velocity is v and the charge density is . at any point between the electrodes at a distance x from the cathode. If I is the equivalent current, m is the mass of each charge carrier, then (A) md v . 2eVx (B) 2eVxA 2 . . md (C) 2md

v . eVx (D) eVxA 2 . . 2md 25. A long round conductor of cross-sectional area A is made of a material whose resistivity depends on the radial distance r from the axis of the conductor as 2 r. . . , . is a constant. The total resistance per unit length of the conductor is R and the electric field strength in the conductor due to which a current I flow in it is . . (A) 2 A R 2 .. . (B) 2 A R 4 .. . (C) 2 A 2 .. I . . (D) 2 A 4..I . .

ELECTRIC CURRENT www.physicsashok.in 45 26. A 100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V source as shown in figure. NowW1, W2 and W3 are the output powers of the bulbs B1, B2 and B3 respectively. Then (A) W1 > W2 = W3 (B) W1 > W2 > W3 (C) W1 < W2 = W3 (D) W1 < W2 < W3 [JEE� 2002 (Scr)] 27. Variation of current passing through a conductor as the voltage applied across its ends is varied as shown in Fig. 9.55. If the resistance is determined at points A, B, C and D. We will find that : (A) Resistance at C and D are equal (B) Resistance at B is higher than at A (C) Resistance at C is higher than at B (D) Resistance at A is lower than at B 28. The P. D. between the points A and B in the circuit shown here is 16 V. Which is/are the correct statement(s) out of the following ? (A) the current through the 2 . resistor is 3.5 A (B) the current through the 4 . resistor is 2.5 A 4. 1. 1. 3. 2. A B 9V 3V (C) the current through the 3 . resistor is 1.5 A (D) the P.D. between the terminals of the 9 V battery is 7 V. 29. A and B are two points on a uniform ring of resistance R. The .ACB = ., where C is the centre of the ring. The equivalent resistance between A and B is (A) . . . . . (2 ) 4R2 (B) . .. . . .. . . . . 2 R 1 (C) R . . 2 (D) R . . . . 4 2 30. The charge flowing through a resistance R varies with time t as Q = at � bt2. The total heat produced in R is (A) 6ba3R (B) 3ba3R (C) 2ba3R (D) ba3 R 31. A resistance R carries a current I . The heat loss to the surroundings is . (T�T0) where . is a constant, T

is the temperature of the resistance, and T0 is the temperature of the atmosphere. If the coefficient of linear expansion is ., the strain in the resistance is (A) proportional to the length of the resistance wire (B) equal to .. I2 R (C) equal to 21 .. I2 R (D) equal to . . (IR) 32. The potential of the point O is (A) 3.5 V (B) 6.5 V (C) 3 V (D) 6 V 33. In the circuit shown in Figure the heat produced in the 5 ohm resistor due to the current flowing through it is 10 calories per second. The heat generated in the 4 ohms resistor is 4. 6. (A) 1 calorie/sec (B) 2 calories/sec 5. (C) 3 calories/sec (D) calories/sec [JEE� 1981]

ELECTRIC CURRENT www.physicsashok.in 46 34. A battery of internal resistance 4. is connected to the network of resistance as shown. In order that the maximum power can be delivered to the network, the value of R in . should be [JEE�1995] (A) 49 (B) 2 (C) 83 (D) 18 35. A steady current flows in a metallic conductor of non-uniform cross-section. The quantity/quantities constant along the length of the conductor is/are [JEE�1997] (A) current, electric field and drift speed. (B) drift speed only (C) current and drift speed (D) current only. 36. In the circuit P . R, the reading of the galvanometer is same with switch S open or closed. Then [JEE�1999] (A) R G . . . (B) P G . . . (C) Q G . . . (D) Q R . . . 37. A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by .T in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of the wire is raised by the same amount .T in the same time t, the value of N is [JEE�2001] (A) 4 (B) 6 (C) 8 (D) 9 38. The effective resistance between points P and Q of the electrical circuit shown in the figure is [JEE�2002] (A) 2Rr R . r (B) 8 . . 3 R R r R r . . (C) 2 r + 4 R (D) 5 2 2R . r 39. Express which of the following set ups can be used to verify Ohm�s law? (A) A V (B) A V (C) A V (D) A V 40. The three resistance of equal value are arranged in the different combination shown below. Arrange them in increasing order of power dissipation. [JEE�2003] (I) a (II) b (III) c (IV) c (A) i (B) i (C) i (D) i R R R

R R 4R E 6R 4. P Q R S G v P Q r r 2R2R 2R 2R2R 2R

ELECTRIC CURRENT www.physicsashok.in 47 41. In the givencircuit, no current is passing through the galvanometer. If the crosssectional diameter ofAB is doubled then for nullpoint ofgalvanometer the value ofACwould [JEE�2003 (Scr)] G A x (A) x (B) x/2 (C) 2x (D) None C B 42. Six equal resitances are connected between points P, Qand R as shown in the figure. Thenthe net resistancewillbemaximumbetween [JEE�2004 (Scr)] P Q R (A) P and Q (B) Q and R (C) P and R (D) any two points 43. For the post office box arrangement to determine the value of unknown resistance, the unknown resistance should be connected between [JEE�2004 (Scr)] B1 C1 B C D (A) B and C (B) C and D (C) Aand D (D) B1 and C1 44. A capacitor is charged using an external battery with a resistance x in series. The dashed line shows the variation of .n . with respect to time. If the resistance is changed to 2x, the new graph will be [JEE�2004] (A) P (B) Q (C) R (D) S 45. A moving coil galvanometer of resistance 100. is used as an ammeter using a resistance 0.1.. Themaximum deflection current in the galvanometer is 100.A. Find the minimum current in the circuit so that the ammeter shows maximum deflection. [JEE�2005] (A) 100.1 mA (B) 1000.1 mA (C) 10.01 mA (D) 1.01 mA 46. In the figure shownthe current through 2. resistor is (A) 2 A (B) 0 A 10V 20V 10 2 5 (C) 4 A (D) 6 A [JEE�2005 (Scr)] 47. Agalvanometer has resistance 100 ..and it requires current 100 µAfor full scale deflection.Aresistor 0.1 . is connectedtomake it anammeter. the smallest current required inthe circuit to produce the fullscale deflection is [JEE�2005 (Scr)] (A) 1000.1 mA (B) 1.1 mA (C) 10.1 mA (D) 100.1 mA 48. Consider a cylindricalelement as shown in the figure. Current flowing through the element is I and resistivityofmaterialof the cylinder is .. Choose the correct option out the following. 4r I l/2 l/2 2r A B C (A) Power loss in second half is four times the power loss in first half. (B) Voltage drop in first half is twice of voltae drop in second half. (C) Current density in both halves are equal.

(D) Electric field in both halves is equal. [JEE�2006]PQRS t ln I

ELECTRIC CURRENT www.physicsashok.in 48 49. Aresistance of 2 . is connected across one gap of a meter-bridge (the length of thewire is 100 cm) and an unknown resistance, greater than 2., is connected across the other gap. When these resistances are interchanged, the balance point shifts by20 cm. Neglecting any corrections, the unknown resistance is [JEE�2007] (A) 3 . (B) 4 . (C) 5 . (D) 6 . MORE THAN ONE CHOICE MAY BE CORRECT 50. Capacitor C1 of capacitance 1micro-farad and capacitor C2 of capacitance 2microfarad are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time t = 0. [JEE�1989] (A) The current in each of the two discharging circuits is zero at t = 0. (B) The currents in the two discharging circuits at t = 0 are equal but not zero. (C) The currents in the two discharging circuits at t = 0 are unequal. (D) Capacitor C1, losses 50% of its initial charge sooner than C2 loses 50% of its initial charge. 51. In the circuit shown in Figure the current through (A) the 3. resistor is 0.50 A. (B) the 3. resistor is 0.25 A (C) the 4. resistor is 0.50 A (D) the 4. resistor is 0.25 A 52. Which of the following is/are wrong? (A) A current carrying conductor is electrically charged. (B) There is an electric field inside a current carrying conductor. (C) For manual control of the current of a circuit, two rheostats in series are preferable to a single rheostat. (D) None of these 53. A constant voltage is applied between the two ensd of a uniform metallic wire. Some heat is produced in it. The heat developed is doubled if: (A) both the length and radius of the wire are halved. (B) both the length and radius of the wire are doubled. (C) the radius of the wire is doubled. (D) the length of the wire is doubled and the radius of the wire is halved. 54. All the resistance in the given Wheatstone bridge have different values and the current through the galvanometer is zero. The current through the galvanometer will still be zero if, (A) the emf of the battery is doubled. (B) all resistance are doubled. (C) resistance R1 and R2 are interchanged. (D) the battery and the galvanometer are interchanged. 55. Two cells of equal emf and having different internal resistance R1 and R2 (R2 > R1) are connected in series. If resistance of connecting wires is equal to R, which of the following statements is/are correct? (A) At a particular value of r, potential difference across second cell can be equal to zero. (B) If R = 0, negative terminal of second cell will be at higher potential than its positive terminal (C) Negative terminal of first cell can never be at higher potential than its positive terminal. (D) None of these 56. When electric bulbs of same power but with different marked voltages are connected in series across a power line, their brightness will be:

(A) proportional to their marked voltages. (B) inversely proportional to their marked voltages. (C) proportional to the squares of their marked voltages. (D) inversely proportional to the square of their voltages. 57. Two conductors made of the same material have lengths L and 2 L, but have equal resistances. The two are connected in series in a circuit in which current is flowing. Which of the following is/are correct? (A) The potential difference across the two conductors is the same. (B) The electric drift velocity is larger in the conductor of length 2L. (C) The electric field n the first conductor is twice that in the second. (D) The electric field in the second conductor is twice that in the first. 2. 2. 2. 2. 2. 4. 3. 9V 8. 8. R1 R2 R3 R4

ELECTRIC CURRENT www.physicsashok.in 49 58. Mark correct statements: (A) Heat is always generated in a battery whether it charges or discharges. (B) When a battery supplies current in an external circuit, heat generated in it is always less than electrical power developed in it. (C) Potential difference across terminals of a battery is always less than its emf. (D)None of these. 59. A mocrometer has a resistance of 100. and a full scale range of 50. A. It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combinations. (A) 50 V range with 10 k . resistance in series. (B) 10 V range with 200 k . resistance in series. (C) 5 mA range with 1. resistance in parallel. (D) 10 mA range with 1. resistance in parallel. 60. An electric current is passed through a circuit containing three wires arranged in parallel. If the length and radius of the wires are in ratio 2 : 3 : 4 and 3 : 4 : 5, then the ratio of current passing through wires would be: (A) 54 : 64 : 75 (B) 9 : 16 : 25 (C) 4 : 9 : 25 (D) 3 : 6 : 10. 61. When a galvanometer is shunted with a 4. resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2. wire, the further reduction in the deflection will be (the main current remains the same). (A) 8/15 of the deflection when shunted with 4. only. (B) 5/13 of the deflection when shunted with 4. only. (C) 3/4 of the deflection when shunted with 4. only. (D) None of these 62. During the charging of a storage battery, the current is 22 A and the voltage is 12 V. The rate of heat generated in the battery is 12 W. The rate of change of internal energy is: (A) 240 J/s (B) 252 J/s (C) 264 J/s (D) 126 J/s. 63. A cell of emf 5 V, internal resistance 1. will give maximum power output to: (A) a single resistor of 1. (B) two 1. resistors connected in series. (C) two 1. resistors connected in parallel. (D) two 2. resistors connected in parallel. 64. In the circuit shown in figure, the current through: (A) the 3. resistor is 0.50 A. (B) the 3. resistor is 0.25 A. 3. 2. 2. 2. 2. 2.(C) the 4. resistor is 0.50 A. (D) the 4. resistor is 0.25 A. 9V 8. 8. 4. 65. For the circuit shown in the figure 6k 1.5k 2k R1 R2 24V RL I [JEE�2009] (A) the current I through the battery is 7.5 mA (B) the potential difference across RL is 18 V (C) ratio of powers dissipated in R1 and R2 is 3 (D) if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9

ELECTRIC CURRENT www.physicsashok.in 50 FILL IN THE BLANKS 1. An electric bulb rated for 500 watts at 100 volts is used in a circuit having a 200 volts supply. The resistance R that must be put in series with the bulb, so that the bulb delivers 500 Watts is ........... ohms. (1997) 2. The equivalent resistance between points A and B of the circuit (Figure) given below is ........... .. 3. In the circuit (Figure) shown above, each battery is 5 V and has an internal resistance of 0.2 ohm. A 2R 2R R B V Fig - 1 Fig - 2 The reading in the ideal voltmeter V is ......... V. [JEE� 1997] TRUE / FALSE 4. Electrons in a conductor have no motion in the absence of a potential difference across it. [JEE� 1982] 5. The current-voltage graphs for a given metallic wire at two different temperatures T1 and T2 are shown in the figure. [JEE� 1985] The temperature T2 is greater than T1. TABLE MATCH 6. Column I Column II (A) Current (P) Mircoscopic quantity (B) Current Density (Q) Macroscopicquantity (C) Electric field (R) Parallel to the conductor boundaries (D) Resistance (S) Flux associated with current density 7. Two bulbs A and B consume same power P when operated at voltage VA and VB respectively. Bulbs are connected with a supply of d.c source then: Column I Column II (A) In series connection, the ratio of (P) RA/RB potential difference across A and B (B) In series connection, the ratio of (Q) V2A/V2B power consumed by A and B (C) In parallel connection, the ratio of (R) RB/RA current in A and B (D) In parallel connection, the ratio of (S) V2B/V2A power consumed in A and B 8. In a R-C circuit. Column I Column II (A) Charging current at tiem t = 0 (P) 1 2 2 CV (B) Discharging current at t = 0 (Q) Maximum (C) While charging energy stored (R) Capacitor becomes short circuit (D) While charging energy dissipated as heat (S) Exponential law VT1 T2 I

ELECTRIC CURRENT www.physicsashok.in 51 9. A battery has an emf E and internal resistance r. A variable resistor R is connected across the terminals of the battery. Column I Column II (A) Current in the circuit is maximum (P) R . . (B) Potential difference across the (Q) R = 0 terminals is maximum (C) Power delivered to the resistor is maximum (R) i Er . (D) Power delivered to the load is zero (S) r = R 10. Consider two identical cells each of emf E and internal resistance r connected to a load resistance R. Column I Column II (A) For maximum power transfer to load (P) 2 4Er if cells are connected in series (B) For maximum power transfer to load if (Q) 2 2Er cells are connected in parallel (C) For series combination of cells (R) eq E . E , eq 2r . r (D) For parallel connection of cells (S) 2 eq E . E , 2 eq r . r PASSAGE TYPE QUESTIONS PASSAGE # I A physics instructor devises a simple electrical circuit setup in which one can easily insert various resistors and capacitors in series and parallel combinations. One can have only resistor combinations, only capacitor combinations, or capacitor-resistor combinations. The circuit is usually used for DC (direct current studies but can also be used for AC (alternating current) studies. The DC battery voltage is 6 volts. The AC rms voltage is 120 volts (at 60 Hz.) The student inserts the resistors and /or capacitors as instructed and has available suitable ammeters and voltmeters for both DC and AC experiments. (There are also three resistors, each of 2 ohms resistance. There are also three capacitors, each of 1 microfarad capacitance.) 1. All three resistors are connected in series and the combination is connected to the 6-volt DC battery. What voltage drop occurs across each individual resistor as measured by the voltmeter? (A) 0.33 V (B) 1.0 V (C) 2.0 V (D) 6.0 V 2. One capacitor and one resistor are connected in parallel. The ends of this combination are then connected t the 6-V DC battery, what are the final current and voltage, respectively, across the 1 microfarad capacitor? (A) 0 A, 6 V (B) 0.33 A, 3 V (C) 0.33 A, 6 V (D) 6 A, 6 V 3. Two of the 2-ohm resistors are connected in parallel and the 120-V AC voltage

is applied to the ends of this parallel combination. What current will the AC ammeter measure if connected so it measures only the current through one of the resistors? (A) 1 A (B) 2 A (C) 60 A (D) 120 A 4. Two resistors are connected in parallel and the set of parallel resistors is then connected in series with the third resistor. If this series parallel combination is connected across the 6-volt battery. What total DC current is drawn from the battery? (A) 0.5 A (B) 2 A (C) 3 A (D) 6 A 5. One capacitor and one resistor are connected is series. The 120 V, 60 Hz, A.C. voltage is then applied to this series �RC� circuit. What is the AC current through this series circuit? (A) 0.045 A (B) 0.50 A (C) 0.72 A (D) 40.0 A

(A) 0.9 W (B) 2.0 W (C) 4.4 W (D) 18 W A 6-votl battery is connected across a 2-ohm resistor. What is the heat energy dissipated in the resistor in 5 minutes? (A) 430 joules (B) 560 joules (C) 4300 joules (D) 5400 joules minutes is used to heat 2 kg of water (which is thermally insulated so that no heat escapes). The initial (A) 0.9 W (B) 2.0 W (C) 4.4 W (D) 18 W A 6-votl battery is connected across a 2-ohm resistor. What is the heat energy dissipated in the resistor in 5 minutes? (A) 430 joules (B) 560 joules (C) 4300 joules (D) 5400 joules minutes is used to heat 2 kg of water (which is thermally insulated so that no heat escapes). The initial ELECTRIC CURRENT PASSAGE # II

A set of experiments in the physics lab is designed to develop understanding of simple electrical circuit principles for direct current circuits. The student is given a variety of batteries, resistors, and DC meters; and is directed to wire series and parallel combinations of resistors and batteries making measurements of the currents and voltage drops using the ammeters and voltmeters. The student calculates expected current and voltage values using Ohm�s law and Kirchoff�s circuit rules and then checks the results with the meters.

6. A student connects a 6-volt battery and a 12-volt battery in series and then connects this combination across a 10-ohm resistor. What is the current in the resistor? (A) 0.8 A (B) 0.9 A (C) 1.8 A (D) 3.6 A 7. Resistorsof 4ohmsand8ohmsareconnectedinseries.Abatteryof 6voltsisconnectedacrosstheseries combination. How much power, in watts, is consumed in the 8-ohm resistor? (D) 24 W (A) 0.67 W (B) 2W (C) 12 W 8. Two 4-ohm resistors are connected in series and this pair is connected in parallel with an 8-ohm resistor. A 12 volt battery is connected across the ends of this parallel set. What power, in watts, is consumed in the 8-ohm resistor in this case? 9. 10. A 12-volt battery is connected across a 4-ohm resistor and the heat energy dissipated in the resistor in 10 temperature of the water is 20°C and the specific heat of the water is 4184 joule/kg-°C. What is the final temperature of the water at the end of the 10 minutes of heating? (A) 22.6 °C (B) 28.4°C (C) 34.2°C (D) 56.4°C PASSAGE # III In the laboratory, the voltage across a particular circuit element can be measured by a voltmeter. A voltmeter has a very high resistance and should be connected in parallel to the circuit element whose voltage is being measured. Connected improperly, the voltmeter will affect the circuit, interrupt

ing it and preventing current from flowing through the circuit element that it is meant to measure. An experiment is conducted in which a voltmeter is used to investigate voltages in a circuit containing a capacitor and a light bulb. The bulb and the capacitor are connected in series with a battery and the voltmeter is placed in different position: in the first case across the capacitor in the second case across the light bulb, and in the third case across the battery (see figure 1). The voltmeter reading is recorded every 10 seconds. The voltage for Case 1 as a function of time is shown in figure 2. Vvoltmeter capacitor bulb battery Case 1 V voltmeter capacitor bulb battery Case 2 Vvoltmeter capacitor bulb battery Case 3 Voltage Time Fig. 2

A capacitor consists of two conducting plates separated by a nonconducting material. When a battery is connected to a circuit containing a capacitor and a light bulb in series, a current will flow, causing positive charge to accumulate on one capacitor plate and an equal amount of negative charge to accumulate on the other. After the current has flowed for a finite time, the capacitor will be fully charged. The ratio of the absolute amount of charge on one plate to the voltage across the plates is defined as the capacitance; this is constant for a given capacitor. A light bulb is a resistor that, when enough current flows through it, becomes hot enough to emit energy in the form of light. (Note : Assume that the battery has no internal resistance and that the resistance of the light bulb is constant).

www.physicsashok.in 52

ELECTRIC CURRENT ELECTRIC CURRENT 11. In the circuit shows in figure 1, which of the following conditions would indicate that the capacitor was fully charged � I. A voltmeter connected across the capacitor reads a constant voltage. II. The light bulb in the circuit stops shining. III. The voltage across the bulb equals the voltage across the battery (A) I Only (B) III Only (C) I and II only (D) I, II, and III 12. Which one of the following graphs could correctly represent the voltage across the battery as a function of time during the experiment described in the passage � Time (D) Time Voltage Voltage Voltage Voltage After the experiment described in the passage is completed, the battery is taken out of the circuit and the wires are reconnected. Which of the following graphs, represents the voltage across the capacitor as a (A) (B) (C) Time

Time

13. function of time �

(A) Time Voltage (B) Time Voltage (C) Time Voltage (D) Time Voltage 14. How will the voltage across the light bulb vary with time as the capacitor is charging � (A) It will decrease, because as the capacitor plates fill will charge, they will impede further charge, which will decrease the current and the voltage across the bulb. (B) It will remain the same, because as the capacitor plates fill with charge and impede the current, the voltage output of the battery will increase to keep the current constant. (C) It will increase, because as the capacitor plates fill with charge, they will induce further charge, which will create a greater voltage across the bulb. (D) It will increase, because as the capacitor plates fill with charge, the volt

age across the capacitor will decrease, and therefore the voltage across the light bulb will increase. 15. The light bulb shown in figure 1 is replaced first with two identical resistor in series, and then with the same two resistors in parallel. The total time taken for the capacitor to charge is measured in both cases, and found to be longer for the first case. It can be deduced that � (A) When the resistance of the circuit si increased, the capacitance of the capacitor increases. (B) the presence of resistors affects the final voltage across the capacitor plates. (C) the more charge is absorbed by the resistors as the resistance of the circuit increases. (D) the presence of resistors hinders the flow of charge, thus reducing the current in the circuit. 16. withoutremovingthevoltmeter� Inthediagram below, avoltmeterisconnectedinseriestoacircuit that includesabatteryandtwobulbsin series. The bulbs, which had been shining in the absence of the voltmeter immediately stop shining. How might the circuit be modified in order to make the bulbs shine steadily again with their former brilliance voltmeter

bulb

V bulb

battery

(A) V (B) V (C) V (D) V www.physicsashok.in 53

ELECTRIC CURRENT www.physicsashok.in 54 LEVEL # 2 1. In the figure shown, 2. 2. 3. 3. 1 .F 3 .F 6V, 0.5 . find the charge on 2 .F each capacitor in steady state. [in .C ]. 2. In the given circuit diaram, 1. 4. 2. 3. AB CD EF 50 V find the current passing through wire CD [in Ampere] 3. It is required to send a current of 8 A through a circuit whose resistance is 5 . . What is the least number of cells which must be used for their purpose and how should they be connected? Emf of each cell is 2V and internal resistance is 0.5 . . 4. In the circuit shown, the capacitor is charged by a battery of emf 100 V and 1. internal resistance by closing the switch. Calculate the heat generated across 99 . resistance during the charging of 0.1 F 100 V, 1. 99 . s capacitor. [in Joule]. 5. In a Wheatstone�s bridge a battery of 2 volt and internal resistance 2 ohm is used. Find the value of the current through the galvanometer in that unbalanced condition of the bridge when P = 1 ohm, Q = 2 ohm, S = 30 ohm and resistance of galvanometer is 4 ohm. 6. A 20 volt battery with an internal resistance of 6 . is connected to a resistor of x ohms. If an additional 6 . resistance is connected across the battery find the value of x so that external power supplied by battery remains the same. 7. Find how the voltage across the capacitor C varies with time t after capacitors C varies with time t after the shorting of the switch S at C S R R E the moment t = 0. 8. A homogeneous poorly conducting medium of resistivity . fills up the space between two thin coaxial ideally conducting cylinders. The radii of the cylinders are equal to a and b, (a < b) the length of each

cylinder is . . Neglecting the edger effects. Find the resistance of the medium between cylinders.

ELECTRIC CURRENT www.physicsashok.in 55 9. Find the current flowing through the resistance R in the circuit shown in figure. The internal resistance of the batteries are negligible. 10. A circuit shown in figure has resistance R1 = 20. and R2 = 30.. At what value of the resistance Rx will the thermal power generated in it be practically independent of small variation of that resistance ? The AB R1 Rx R2 voltage between the points A and B is supposed to be constant in this case. 11. An ammeter and voltmeter are connected in series to a battery with an emf E = 6.0v. When a certain resistance is connected in parallel with the voltmeter, the reading of the latter decreases . = 2.0 times whereas the reading of ammeter increases the some number of time. Find the voltameter reading after the connecting of the resistance. 12. In the circuit shown in the figure the current through 3 . resistance is 2A. If E1 = 12V, E2 = 14V, what is the value E1 E2 E3 3. 1. 1.5. 10. 1. of E ? Internal resistance of each battery is 1 . . 13. The circuit shows a capacitor C two batteries, two resistors and a switch S. Initially S has been open for a longtime. It is then closed for a long time by how much does the change on the capacitor change over this time period ? Assume C = 10 ., E1 = 1.0 V, E3 = 3.0v, R1 = 0.20 . , R2 = 0.40 . . 14. In the given network switch S is closed at t = 0. Find the current through the 10 ohm resistor at t = 75 . sec. 10V 20V S 2.F 10. .. .. 15. Two coils connected in series have resistance of 600 . and 300 . and temperature co-efficient of 0.001 and 0.004 (°C)�1 respectively at 20°C. Find resistance of the combination at a temperature of 50°C. What is the effective temperature co-efficient of combination. 16. The resistance of the galvanometer G in the circuit is 25.. The meter deflects full scale for a current of 10 mA. The meter behaves as an ammeter of three different ranges. The range is 0�10 A, if the terminals R1 R2 R3 G O and P are taken; range is 0�1 A between O and R. Calculate the resistance R1, R2 and R3.

ELECTRIC CURRENT www.physicsashok.in 56 17. Calculate the steady current in the 2-ohm resistor shown in the circuit in the figure. The internal resistance of the battery is negligible and the capacitance of the conductor C is 0.2 microfarad. OR Two resistors, 400 ohms, and 800 ohms are connected in series with a 6-volt battery. It is desired to measure the current in the circuit. An ammeter of a 10 ohms resistance is used for this purpose. what will be the reading in the ammeter? similarly, If a voltmeter of 10,000 ohms resistance is used to measure the potential difference across the 400-ohms resistor, What will-be the reading in the voltmeter. LEVEL # 3 1. A part of circuit in a steady state along with the currents flowing in the branches, the values of resistances etc., is shown in the figure. Calculate the energy stored in the capacitor C (4.F) [JEE� 1986] 2. An infinite ladder network of resistances is constructed with a 1 ohm and 2 ohm resistances, as shown in figure. [JEE� 1987] The 6 volt battery between A and B has negligible internal resistance: (i) Show that the effective resistance between A and B is 2 ohms. (ii) What is the current that passes through the 2 ohm resistance nearest to the battery? 3. An electrical circuit is shown in Figure. Calculate the potential difference across the resistor of 400 ohm, as will be measured by the voltmeter V os resistance 400 ohm, either by applying Kirchhoff�s rules or otherwise. [JEE� 1996]

ELECTRIC CURRENT www.physicsashok.in 57 4. Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs V1 and V2 and internal resistance r1 and r2 respectively, with polarities as shown in figure. [JEE� 1997] 5. A leaky parallel plane capacitor is filled completely with a material having dielectric constant K = 5 and electrical conductivity . = 7.4 x 10�12 .�1 m�1. If the charge on the plane at instant t = 0 is q = 8.85 mC, then calculate the leakage current at the instant t = 12 s. [JEE� 1997] 6. In the circuit shown in Figure, the battery is an ideal one, with emf V. The capacitor is initially uncharged. The switch S is closed at time t = 0. [JEE� 1998] (A) Find the charge Q on the capacitor at time t. (B) Find the current in AB at time t. What is its limiting value as t ... : 7. A thin uniform wire AB of length 1m, an unknown resistance X and a resistance of 12. are connected by thick conducting strips, as shown in the figure. A battery and a galvanometer (with a sliding A B C D jockey connected to it) are also available. Connections are to be X 12. made to measure the unknown resistance X using the principle of Wheatstone bridge. Answer the following questions. [JEE� 2002] (A) Are there positive and negative terminals on the galvanometer? (B) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at approxiate points. (C) After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. Obtain the value of the resistance of X. 8. How a battery is to be connected so that the shown rheostat will behave like a potential divider? Also indicate the points out A B C R which output can be taken. [JEE� 2003] 9. Draw the circuit diagram to verify Ohm�s Law with the help of amain resistance of 100. and two galvanometers of resistances 106 . and 10�3 . and a source of varying emf. Show the correct positions of voltmeter and ammeter. [JEE� 2004] 10. An unknown resistance is to be determined using resistances R1, R2 or R3. Their corresponding null points are A, B and C. Find which of the above will give the most accurate reading and why? [JEE� 2005] A B C G X R R = R1 or R2 or R3

ELECTRIC CURRENT www.physicsashok.in 58 ANSWER KEY Reasoning Type Que. 1 2 3 4 5 6 7 8 9 10 11 12 Ans. C C D A C A A C A D B A Level # 1 Q. 1 2 3 4 5 6 7 8 9 10 Ans. C B D D A C C D D B Q. 11 12 13 14 15 16 17 18 19 20 Ans. B B B D CD A A ABCD A C Q. 21 22 23 24 25 26 27 28 29 30 Ans. B C B AB C D A ACD A A Q. 31 32 33 34 35 36 37 38 39 40 Ans. B A B B D A B A A A Q. 41 42 43 44 45 46 47 48 49 50 Ans. A A C B A B D A A BD Q. 51 52 53 54 55 56 57 58 59 60 Ans. D AC B ABD ABC C AC AB BC A Q. 61 62 63 64 65 Ans. B B AD D AD Fill in the Blanks / True�False / Match Table 1. 20 2. R 2 3. 0 4. F 5. T 6. A .Q, R and S, B .P and R, C .P and R, D .Q 7. A .P and Q, B .P and Q, C .R and S, D .R and S 8. A .Q, R and S, B .Q and S, C .P and S, D .P and S 9. A .Q and R, B .P, C .S, D .Q, P and R 10. A .Q, B .Q, C .S, D .R Que. 1 2 3 4 5 6 7 8 9 10 Ans. C A C B A C B D D A Que. 11 12 13 14 15 16 Ans. C A B A D A Passage Type

ELECTRIC CURRENT www.physicsashok.in 59 Level # 2 1. 72µC 11 2. 2 A 3. 160 4. 495 J 5. 19 Amp 6. x = 7.5. 7. V = . 2t RC. E 1 e 2 . . 8. ... ... .. an b 2 . . 9. I = . . . . . . 2 3 2 32 3 0 3 R R R R RE R R E R . .. . 10. Rx = 1 2 1 2 R R R R . = 12. 11. 1 E. . = 2.0 V 12. E = 7V 13. decreases by 13.3 .c 14. I = 20 mA 15. 954 . , 0.002 (°C)�1 16. R1 = 0.025 . , R2 = 0.2275 . , R3 = 2.5275 . 17. 0.9A or 4.96 x 10�3 A, 1.95 V Level # 3 1. 2.88 x 10�4 J 2. (ii) 1.5 A 3. 10/3 V 4. 1 2 2 1 1 2 V r V r r r .. , 1 2 1 2 r r r . r 5. 0.199.A 6. (a) . . 1 2 3 2 CV . e. t RC ; (b) 2 3 2 6 V V e t RC R R . . ; 2

VR 7. (a) No (b) A B C D X 12. G J (c) 8. 8. Battery should be connected across A and B. Out put can be taken across the terminals A and C or B and C. 9. G1 G2 103 Ammeter Voltmeter 106 100E 10. This is true for r1 = r2; So R2 given most accurate value. �X�X�X�X�

MAGNETIC FIELD

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

MAGNETIC FIELD www.physicsashok.in 1 THEORY OF MAGNETICEFFECTOFELECTRICCURRENT CONCEPT OF MAGNETIC FIELD The space around a current carrying conductor, inwhich itsmagnetic effect can be experienced, is called magnetic field. Inmagnetics, there are basicallytwomethods ofcalculatingmagnetic field at some point.One isBiot Savert�s lawwhich gives themagnetic field due to an infinitesimallysmall current carryingwire at some point and the other isAmpere�s law, which is useful in calculating themagnetic field of a highly symmetric configuration carrying a steadycurrent. BIOT-SAVART�S LAW According to this law, the magnetic field dB.. at the point P due to the small current element of length d.lis given by 0 2 2 dB µ id sin Wb /m or tesla 4 r . . . l where µ0 is a constant and is called, permeabilityof free space. i r P µ0 = 4. × 10�7 Wb/A�m Rules to Find the Direction of Magnetic Field (i) Right hand palm rule no. 1 : Ifwe spread our right hand in such a way that thumbis towards the directionof current and fingures are towards that point wherewe have to find the direction of field then the direction of field i Current carrying conductor P B will be perpendicular to the palm (ii) Maxwell�s right handed screw rule : If a right handed cork screw is rotated so that its tip moves in the direction offlowofcurrent throughthe conductor, then the rotationof the head ofthe screwgives the direction ofmagnetic lines of force. × P2 P1 Magnetic line of force i Current carrying conductor NOTE : By convention the direction of magnetic field ..Bperpendicular to paper going inwards is shown by . and the direction perpendicular to the paper coming out is shown by . . Applications of Biot-Savart�s Law

Let us consider fewapplications ofBiot-Savart�s Law: (i) Magnetic field due to a straight thin conductor is 0 . . 1 2 B µ i sin sin 4 d . . . . . d 12 p i (a) For aninfinitelylong straight wire, .1 = .2 = 90º

MAGNETIC FIELD www.physicsashok.in 2 . B µ0i 2 d . . (b)Whenwire is semi-infinite, 1 2 0 and 2. . . . . i . d 0 µ i B sin 0 sin 4 d 2. .. . . . . . . . 0 µ i B 4 d . . (c) B 1d . , i.e., B�d graph for aninfinitely long straight wire is a rectangular hyperbola as shown in figure. B (ii) Magnetic field on the axis ofa circular coil havingNturns is d . . 2 0 2 2 3/ 2 B µ NiR 2 R x . . Here, R = radius of the coil O x R P x = the distance of point P fromcentre and i = current in the coil (a)At the centre of the loop, x = 0 . 0 B µ Ni 2R . (b) For x > > R, x2 + R2 . x2 . 2 2 0 0 0 3 3 3 µ NiR µ 2Ni R µ 2M B 2x 4 x 4 x . . . . . . . . . . . . . . . . . . . . . .. . . . .. . . Here, M=magneticmoment of the loop M = NiA = Ni .R2 NOTE : This result was expected since, the magnetic field on the axis of dipole is 0 3 µ 2M

4p x Example 1. Two circular coilsAand Bof radius 52 cmand 5 cmrespectively carry current 5Amp and 52 Amp respectively.The plane ofB is perpendicular to plane ofAand their centres coincide. Find themagnetic field at the centre. Sol. Themagnetic field due to first coil is 7 0 1 1 1 2 µ I 4 10 5 B 5 2r 2 10 2 . . . . . . . . .7 1 2 20 3.14 10 B 5 1.441 10. . . . . . . B1 = 8.88 × 10�5 web/m2

MAGNETIC FIELD www.physicsashok.in 3 and 7 0 2 2 2 2 µ I 4 10 5 B 2r 2 5 10 2 .. . . . . . . . . 5 5 2 2 3.14 10 B 4.44 10 1.414 . . . . . . . . 2 2 B . B1 . B2 B . (8.88)2 . (4.44)2 .10.5weber /m2 B . 78.85 . 19.71 . 9.93 wb / m2 Example 2. Three rings, each having equal radius R, are placedmutuallyperpendicular to each other and each having its centre at the origin of co-ordinate system. If current �I� is flowing through each ring then themagnitude ofthemagnetic field at the common centre is (A) 0 3 µ I 2R (B) zero (C) . . 0 2 1 µ I 2R . (D) . . 0 3 2 µ I 2R . Sol. Themagnetic field due to ring inx�yplane is 0 1 B µ I k� 2R . .. themagnetic field due to ring iny�z plane is 0 2 B µ I �i 2R . .. and themagnetic field due to ring in x�z plane is 0 3 B µ I �j

2R . .. . B . B1 .B2 .B3 .. .. .. .. 0 . . B µ i k� i� �j 2R . . . .. . 0 B 3µ I 2R . Hence, option (A) is correct (c)Magnetic field due to an arc of a circle at the centre is 0 0 µ i µ i or B 2R 4 R . . . . . . . . . . . . . . . . .. . .. . . B= 2 O R Inwards i (iii) Field along the axis of a solenoid is 1 2 R x L O

MAGNETIC FIELD www.physicsashok.in 4 0 . . 2 1 B µ Ni cos cos 2 . . . . (a) For a long solenoid (L >>R), i.e., .1 = 180º and .2 = 0º B = µ0Ni (b)At the ends of solenoid, .2 = 0º, .1 = 90º we get, 0 B 1 µ Ni 2 . (for L>>R) Example 3. In a high tensionwire electric current runs fromeast to west. Find the direction ofmagnetic field at point above and belowthewire. Sol. When the current flows fromeast to west,magnetic field lines are circular round it as shown infigure (a).And so, themagnetic field above thewire is towards north and belowthewire towards south. W NE S B i (a) (b) Example 4.A0.5 mlong solenoid has 500 turns and has a flux density of 2.52 × 10�3 T at its centre. Find the current in the solenoid. Given, µ0 = 4. × 10�7Hm�1. Sol. Here, B = 2.52 × 10�3 T; µ0 = 4. × 10�7 Hm�1 Length of the solenoid, l = 0.5m; Total number ofturns in the solenoid,N= 500 Therefore, number ofturns per unit lengthof the solenoid, n N 500 1000 m 1 0.5 . . . . l If �i� is the current throughthe solenoid, then B=µ0ni or 3 7 0B 2.52 10 i 2.0 A µ n 4 10 1000 . .. . . . . . . Example 5. Arectangular polygon of �n�sides is formed by bending a wire of total length 2.R which carries a current �i�. Find themagnetic field at the centre of the polygon. Sol. One side of the polygon is,

a 2 R n. . 2n2 n . .. .. .. . . . . . . d a i . . d cot a / 2 . . . d a cot R cot 2 n n . . . . . . .. . . . . . . . . . . . . . . .

MAGNETIC FIELD www.physicsashok.in 5 All sides of the polygon produce the magnetic field at the centre in same direction (here . ). Hence, net magnetic field, B = (n) (magnetic field due to one side) 0 . . µ i B n sin sin 4 d . . . . . . . . . . . or 0 µ in B n tan 2sin 4 R n n . . . . . .. . .. . . . . . . . . . . .. . . . . . . . or . . 2 0 sin µ i n n B 2R cos / n . . .. . . . .. .. . . . . . . .. .. . Example 6. Infinite number of straight wires eachcarrying current �I� are equally placed as shown in the figure.Adjacent wires have current in opposite direction. Netmagntic field at point P is 30º 1 2 3 4 5 y x z a a P 30º (A) 0 µ I n 2 �k 4. 3 a l (B) 0 µ I n 4 �k 4. 3 a l (C) 0 µ I n 4 ( k� ) 4 3 a . . l (D) Zero Sol. . . 1 2 3 4 B . B . B . B . B ........... k� .. Here 0 . . 1 B µ I sin 30º sin 30º 4 a . . . 0 1 µ I 1 1 B 4 a cos30º 2 2 . . . . . .. ..

0 1 2µ I B 4 3 a . . Similarly, 0 2 2µ I B 8 3 a . . and so on . 0 2µ I 1 1 1 B 1 ........ 4 3 a 2 3 4 . . . . . . .. . .. .. .. 0 2µ I � B ln 2k 4 3 a . . .. 0 µ I � B ln 4k 4 3 a . . .. Hence, option (B) is correct. Example 7. Along straight wire, carrying current I, is bent at itsmidpoint to formanangle of 45º.Magnetic field at point P, distanceRfrompoint of bending is equal to : 45º I I P R (A) . . 0 2 1 µ I 4 R .. (B) . . 0 2 1 µ I 4 R .. (C) . . 0 2 1 µ I 4 2 R . . (D) . . 0 2 1 µ I 2 2 R . .

MAGNETIC FIELD www.physicsashok.in 6 Sol. 0 . . B µ I sin 90º sin135º 4 . R2 . . . 0 . . B µ I 2 1 4 R . . . Hence option (A) is correct Example 8. Find themagnetic field at P due to the arrangement shown (A) 0 µ i 1 1 2 d 2 . . . . . . . . . (B) 0 2µ i 2 d . . 90º 45º d P (C) 0 µ i 2 d . . (D) 0 µ i 1 1 2 d 2 . . . . . . . . . Sol. B . B1 . B2 .. .. 0 B µ I sin sin 2 4 d 4 2 2 . . .. . .. . . . . . . . 0 B µ I 1 1 2 d 2 2 . . . . . .. .. . 90º 45ºP45º /4 d2d20 2 µ I 2 1 B 2 d 2 . . . . . . . . . . . 0 . . B µ I 2 1 2 d . . . . 0 B µ i 1 1 2 d 2 . . . . . . .. .. Hence option (A) is correct Example 9.What is themagnitude ofmagnetic field at the centre �O� of loop of radius 2 mmade of uniformwirewhen a current of 1 amp enters in the loop and taken out of it by two longwires as shown in the figure. 45º

1 amp 90º 1 amp O Sol. 1 B .. =magnetic field due to left wire is 0 1 µ I � B sin sin k 4 d 2 4 . . .. . . . . . . . O/4 d 2 B .. =magnetic field due to right wire 0 µ I � sin sin ( k) 4 d 2 4 . . .. . . . . . . . . In circularwire, I 1R . . 12 II 2 . . . . . I1 I2 � . 1 2 I I 2 . . . . . . . . . ..

MAGNETIC FIELD www.physicsashok.in 7 . 3 B .. =magnetic field due to circularwire 0 1 0 2 µ I µ I (2 ) 4 r 4 r . . . . . . . . 0 2 0 2 µ I µ (2 ) I 0 4 r 4 r(2 ) . . . . . . . . . . . . . . B . B1 . B2 . B3 . 0 .. .. .. .. AMPERE�S CIRCUITAL LAW It states that the line integral of B.. around any closed path or circuit is equal to µ0 times the total current crossing the area bounded bythe closed path provided the electric field inside the loop remains constant. Thus, . . 0 net C . B. d . µ i .. .l Its simplified formis Bl = µ0 inet This equation canbe used onlyunder following conditions : (a) at every point of the closed path B|| dI .. . (b)magnetic field has the samemagnitudeB at all places on the closed path. Applications of Ampere�s Circulatal Law (i) Magnetic field due to a longmetal rod of radiusR carrying a current �i� : (a) If r < R, 0 2 µ i B r 2 R . . . .. . .. , i.e., B .r (b) If r = R (i.e., at the surface) 0 B µ i 2 R . . (ii) Magnetic field of a solenoidwounded in the formof a helix is B = µ0Ni NOTE : Ampere�s law is valid only for steady currents. Further more, it is useful only for calculating the magnetic fields of current configurations with high degrees of symmetry, just as gauss�s law is useful only for calculating the electric fields of highly symmetric charge distributions. Example 10. Two long conductors are arranged as shown above to form overlapping cylinders, each of raidus �r�, whose centers are separated by a distance �d�. Current of densityJ flows into the plane of the page along the shaded part of one conductor and an equal current flows out d Conductor

y x Vacuum A of the plane of the page along the shaded portion of the other, as shown. What are themagnitude and direction of themagnetic field at pointA? (A) (µ0/2.).dJ, in the +y-direction (B) (µ0/2.)d2/r, in the +y-direction (C) (µ0/2.)4d2J/r, in the �y-direction (D) (µ0/2.)Jr2/d, inthe �y-direction Sol. B . B1 . B2 .. .. .. 0 0 µ d � µ d � B j i j i 2 2 2 2 . . . . . . . . . . . . . .. . . . . . . . . .. . .

MAGNETIC FIELD www.physicsashok.in 8 B µ0 jk� d �i µ0 jk� d i� 2 2 2 2 . . . . . . . . . .. .. .. .. .. 0 0 B µ jd �j µ jd �j 4 4 . . .. 0 0 B µ jd �j µ d j 2 2. . . . .. along y-axis Hence, option (A) is correct. MAGNETIC FIELD OF A MOVING POINT CHARGE The magnetic field vector B.. at point P at position vector r . , fromthe charge qmovingwith a velocityv . is found to be 0 . . 3 B µ q v r 4 r . . . . .. . . Note downthe following points regarding this equation. (a)Magnitude ofBis, 0 2 B µ qvsin 4 r . . . r vq p It is zero at . = 0º and 180º andmaximumat . = 90º (b) Direction ofB.. is along v r . . . if q is positive and opposite to v . r . . if q is negative. (c) Suppose a charge q1 is moving with velocity and 1 v . another charge q2 is moving with velocity 2 v . at position vector r . relative to q1, then force on q2will be, . . 2 2 F . q v . B . . .. . . 0 1 . . 2 3 1

µ q F v . v r 4 r . . . . . . . . . . . . . . . . . . . . 0 1 2 . . . . 3 2 1 F µ . q q v v r 4 r . . . . . . . . . . . . This corresponds to Coulomb�s electricalforce between the charges q1 and q2movingwith velocities 1 v . and 2 v . respectivelyrelative to an observer at rest. Example 11. For a charge �q�movingwith velocityv . , find the relation between electric andmagnetic fields. Sol. 3 0 E q 4 r . .. .. ...(i) 0 3 B µ qv r 4 r. . ... . . ...(ii) and 0 0 c 1 µ . . or 2 0 0 c 1 µ . .

MAGNETIC FIELD www.physicsashok.in 9 FromEqs. (i) and (ii), we get B . µ0.0 v . E .. . .. . 2 B v E c. . .. . .. FORCE ON A CURRENT CARRYING CONDUCTOR IN MAGNETIC FIELD When a current carrying conductor is placed in amagnetic field, the conductor experiences a force in a directionperpendicular to both the direction ofmagnetic field and the direction of current flowing in the conductor. This force is also called B F i Lorentz force. i The direction of this force can be found out either by Fleming�s left hand rule or by right handpalmrule. Themagnetic force is F = ilB sin . In vector form, F . i. . B. . . .. l Where B= intesityofmagnetic field i= current in the conductor l = lengthof the conductor and . = angle between the length of conductor and directionofmagnetic field. Case : (i) If ..= 90º or sin ..= 1 then F = ilB(maximum) Therefore, forcewillbemaximumwhen the conductor carrying current is perpendicular tomagnetic field. (ii) If ..= 0º or sin ..= 0, Then F = ilB × 0 = 0 Thus, the forcewill be zero,when the current carrying conductor is parallel to the field. Example 12. Astraight current carrying conductor is placed in such away that the current in the conductor flows in the direction out of the plane of the paper.The conductor is placed between two poles of two magnets, as shown. The S N Q R S P conductor willexperience a force inthe direction towards. (A) P (B) Q (C) R (D) S Sol. The direction ofmagnetic field on the conductor is along SR. But F . i . B . . .. l F . i k� . (�B�i) . l F . .i B�j . i B(.�j) .

l l Hence, the direction ofmagnetic force on thewire is towards Q. Hence option (B) is correct Example 13. In the figure shown a semicircularwire loop is placed in uniformmagnetic fieldB = 1.0T.The plane of the loop is perpendicular to themagnetic field. Current i= 2Aflows inthe loop in the direction shown. Find themagnitude of themagnetic force in both the cases (a) and (b). The radius of the loop is 1.0m. i = 2A i = 2A × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × (a) (b) 1m

MAGNETIC FIELD www.physicsashok.in 10 Sol. Refer figure (a) : It forms a closed loop and the current completes the loop. Therefore, net force on the loop in uniformfield should be zero. Refer figure (b) : In this case although it forms a closed loop, but current does not complete the loop.Hence, net force is not zero.FACD . FAD . . . Floop . FACD . FAD . 2 FAD . . . . × × C × × × × × × × × × × × × × × B A D . | Floop | . 2| FAD | . . | Floop | . 2i B sin . . l (l = 2r = 2.0 m) = (2) (2) (2) (1) sin 90º = 8 N Example 14. An arc of a circular loop of radius �R�is kept in the horizontal plane and a constantmagnetic field �B� is applied in the verticaldirection as shown in the figure. If the arc carries current �I� then find the force in the arc. × B 90º I × × × × × × × × × × × × × × × × × × × × × Sol. Aswe know,magnetic force on a closed loop placed in uniformmagnetic field is zero. × × × × × × . B1 . B2 . 0 .. .. . 1 2 B . .B . .I B(.�j) .. .. l 1 B . I B�j .. l /2 R R (1) (2) 1 B . I B( 2R) �j .. . B1 . 2 IBR .. Example 15. Aconducting wire bent in the formof a parabola y2 = 2x carries a current i = 2Aas shown in figure. thiswire is placed in a uniformmagnetic field B . .4 k� .. Tesla. Themagnetic force on thewire is (in newton) y(m) AB

2 x(m) (A) .16�i (B) 32�i (C) .32�i (D) 16 �i Sol. The netmagnetic force on closed loop is zero. . Force on parabola + force on straight wireAB = 0 . Force on the parabola = � force on straight wireAB . ..I�j. B. .. AB . .2..4�j. ..4k� .. F . .32 �i Hence option (C) is correct. Example 16. Aconductor of length �l � and mass �m�is placed along the east-west line on a table. Suddenly a certain amount of charge is passed through it and it is found to jump to a height �h�. The earth�smagnetic induction isB. The charge passed throughthe conductor is : (B is horizontal) (A) 1 Bmgh (B) 2gh Blh (C) gh Blh (D) m 2gh Bl Sol. Themagnetic force is F = I lB or F dq B dt . l

MAGNETIC FIELD www.physicsashok.in 11 or dq Fdt B . l or 0 q Fdt mv B B . . . l l But 02 = v02 � 2gh . 0 v . 2gh . m 2gh q B . . l Hence option (D) is correct Example 17. AU-shapedwire ofmassmand length l is immersedwith its two ends inmercury(see figure).Thewire is ina homogeneous field ofmagnetic inductionB. If a charge, that is, a current pulse q . .idt , is sent through thewire, thewirewill jump up. × × × × × × × × × × × × × × × × × × × × × l m B i Calculate, fromthe height �h� that thewire reaches, the size of the Hg charge or current pulse, assuming that the time of the current pulse is verysmall incomparisionwiththe time offlight.Make use ofthe fact that impulse offorce equals . Fdt ,which equalsmv. Evaluate �q�for B = 0.1Wb/m2, m= 10 gm, l = 20 cm&h = 3 meters. [g = 10 m/s2] Sol. I l B = F . Fdt = mv or I l Bdt = mv But 1 mv2 mgh 2 . . v . 2gh . Idt mv m 2gh B B . . l l . 3 2 q m 2gh 10 10 2 10 3 15 coulmb. B 20 10 0.1 . . . . . . . . l . .

Example 18.Ametal ring of radius r = 0.5mwith its plane normal to a uniformmagnetic fieldB ofinduction 0.2T carries a current I = 100A. The tension in newtons developed in the ring is : (A) 100 (B) 50 (C) 25 (D) 10 Sol. We consider a small portion .l of the loop. For equilibrium, F 2T sin 2T T 2 2 . . . . . . or I.lB = T. ......... Tcos /2 /2 /2 Tcos /2 T T F + B or Ir.B=T. . T = IrB = 100 × 0.5 × 0.2 = 10 N Hence option (D) is correct

MAGNETIC FIELD www.physicsashok.in 12 FORCE ON A MOVING CHARGE IN MAGNETIC FIELD : LORENTZ FORCE If a charged particlemoves in amagnetic field, then a force acting on it is given by F = qvB sin . In vector form, F . q.v . B. . . .. × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × FCases : v (i) If v= 0, then F = 0 i.e., no force is exerted on a stationary charge, in a magnetic field. (ii) If ..= 0, then F = 0 i.e., when the charge ismoving parallel to the field then no forcewillbe exerted bythe field. (iii) If ..= 90º, then sin ..= sin 90º = 1 F = qvB × 1 = qvB i.e., when the charged particle is moving parpendicular to the field, the force exerted by the field will be maximum. Rules of Find the Direction of Force (i) Right hand palm rule : Ifwe stretch the right hand palmsuch that the fingers and the thumb are mutually perpendicular to each other and the fingers point the direction ofmagnetic field and the thumb points the direction ofmotionof positive charge, the direction of forcewillbe along the outward normalon the palm. Field B Force F Current or motion of positive charge (ii) Fleming�s left hand rule : Ifwe spread the forefinger, central finger and thumb of our left hand insuch away that these three are perpendicular to each other then, if first forefinger is in the direction ofmagnetic field, second centralfinger is in the direction of current, then thumbwill represent the direction of force. Current or motion of positive charge Force F Field B v NOTE : To learn this rule, remember the sequence of Father, Mother, child. Thumb . Father . Force Forefinger . Mother . Magnetic field Central finger . Child . Current or direction of positive charge Example 19.Whena protonhas a velocity v . .2�i . 3�j..106 m/ s . it experiences a force F . ..1.28 .10.13 k� .N. . When its velocityis along the z-axis, it experiences a force along the x-axis.What is themagnetic field ? Sol. Substituting proper values in,

F . q.v . B. . . .. We have, . 13 . . 19 . . . . . 6 0 . 1.28.10. k� . 1.6 .10. . 2�i . 3�j . .B �j . .10 . .

MAGNETIC FIELD www.physicsashok.in 13 . 1.28 = 1.6 × 2 × B0 or 0 B 1.28 0.4 3.2 . . Therefore, themagnetic field is, B . ..0.4�j.T .. MOTION OF CHARGED PARTICLE IN MAGNETIC FIELD The pathofchargedparticle inuniformmagnetic field dependsonangle betweenv . andB.. .Therefore, following cases are possible : Case-I .When . is 0º or 180º : The magnetic force is F = Bqv sin 0º or sin 180º = 0. Hence, path of the charged particle is a straight line (undeviated) when it enters parallel or antiparallel to magnetic field. Case-II.When . = 90º : Themagnetic force is F = Bqv sin 90º = Bqv.Thismagnetic force is perpendicular to the velocity at every instant. Hence, path is circle. The necessarycentripetal force is provided by the magnetic force hence, if �r�be the radius of the circle, then + + q v v q or B B Fm= 0 mv2 qBv r . . r mv qB . This expression of �r�can bewritten in following different ways : mv p 2Km 2qVm r qB qB qB qB . . . . Here, P =momentumof particle K= KE of particle p2 or p 2Km 2m . . Further, time period of the circular pathwill be 2 mv 2 r qB 2 m T v v qB . . .. . . . . . . . . or T 2 m qB . . or the angular speed (.) of the particle is 2 qB

T m . . . . . qB m . . Frequencyof rotation is,1f or f qB T 2 m . . .

MAGNETIC FIELD www.physicsashok.in 14 IMPORTANT FEATURES If angle . is other than 0º, 180º or 90º, then velocity of charged particle can be resolved in two components one along B.. and another perpendicular to B.. . Let the two components be | | v and v. . Then v| | . v cos. and v vsin . . . q, m + B v v sin v cos The component perpendicular to field ( v. ) gives a circular path and the component parallelto field ( | | v ) gives a straight line path. The resultant path is a helix as shown in figure. The radius ofthis helicalpath is, r mv mvsin qB qB . . . . Time period and frequency do not depend on velocity and so theyare given by T 2 m and f qB qB 2 m . . . . There is onemore termassociatedwitha helicalpath, that is pitch (p) of the helicalpath. Pitch is defined as the distance travelled alongmagnetic field in one complete cycle, i.e., p = v| |T or p .vcos . 2 m qB . . . . . p 2 mvcos qB . . . Example 20. What is the smallest value of B that can be set up at the equator to permit a proton of speed 107m/s to circulate around the earth ? [R = 6.4 × 106 m, mp = 1.67 × 10�27 kg]. Sol. Fromthe relation R mv qB . We have B mv qR . Substituting thevalues,we have . .. . . .. . 27 7 8 19 6 1.67 10 10

B 1.6 10 T 1.6 10 6.4 10 . . . . . . . . . Example 21. Ablock ofmassm&charge �q�is released on a long smooth inclined planemagnetic field Bis constant, uniform, horizontal and parallelto surface as shown. Find the time fromstart when block loses contact with the surface. B m q (A) mcos qB . (B) m cosec . qB (C) m cot . qB (D) none

MAGNETIC FIELD www.physicsashok.in 15 Sol. For losing the contects N = 0 . qvB=mgcos. . v mg cos qB . . But v = u + at or v = 0 + gt sin . . gt sin mg cos qB . . . . t mcot qB . . Hence option (C) is correct. Example 22.An electronhaving kinetic energyTismoving in a circular orbit ofradiusRperpendicular to a uniform magnetic inductionB.. . If kinetic energyis doubled andmagnetic induction tripled, the radiuswillbecome (A) 3R2 (B) 3R 2 (C) 2R 9 (D) 4R 3 Sol. R mv P qB qB . . But kinetic energy P2 T 2m . . P . 2mT . 2mT R qB . . 2m(2T) R´ q(3B) . R´ 2 R 2 R 3 9 . . Hence, option (C) is correct. Example 23. Acharged particle (charge q, massm) has velocity v0 at origin in +x direction. In space there is a

uniformmagnetic fieldB in �z direction. Find the �y� coordinate of particlewhenis crosses y�axis. Sol. The parth ofcharged particle is circularwhose radius is 0 mv r qB . O C + B . 0 2mv y 2r qB . . Example 24. Amass spectrometer is a devicewhich select particle of equal mass.An ironwith electric charge q > 0 starts at rest froma source S and is accelerated through a potentialdifferenceV. It passes through a hole into a region of constant magnetic field B.. perpendicular to the B V S plane of the paper as shown in the figure. The particle is deflected by themagnetic field and emerges throughthe bottomhole at a distance �d� fromthe top hole. Themass of the particle is (A) qBd V (B) qB2d2 4V (C) qB2d2 8V (D) qBd 2V

MAGNETIC FIELD www.physicsashok.in 16 Sol. The speed of charged particle just before entering themagnetic field isV0. 20qV 1 mv 2 . 0 v 2qV m . The radius of circular path inmagnetic field is 0 mv r qB . or 0 d mv 2 qB . or 2 2 2 2 0 2 2 2 2 d m v m 2qV 4 q B q B m . . or 2 2 d 2mV 4 qB . . qB2d2 m 8V . Hence, option (C) is correct Example 25. Acyclotron is operatingwith a flux density of 3Wb/m2. The ionwhich enters the field is a proton havingmass 1.67 × 10�27 kg. If themaximumradius of the orbit of the particle is 0.5m, find : (a) themaximumvelocityof the proton, (b) the kinetic energy of the particle, and (c) the period for a half cycle. Sol. (a)As in case ofmotion of a charged paricle in amagnetic field, r mv i.e., v qBr qB m . . So, max max v qBr m . So, 19 8 max 27 1.6 10 3 0.5 v 1.43 10 m/ s

1.67 10 . . . . . . . . . (b) . .K 1 mv2 1 1.67 10 27 1.43 108 2 2 2 . . . . . . . i.e. 11 11 19 1.71 10 K 1.71 10 J 107 MeV 1.6 10 . . . . . . . . . (c) In case of circularmotion, as 8 8 T 2 r 2 0.5 2.19 10 s v 1.43 10 . . . . . . . . . So, time for helf cycle, T 1 (T) 1.09 10 8 s 2 . . . .

MAGNETIC FIELD www.physicsashok.in 17 Example 26. The region between x = 0 and x = Lis filledwith uniformsteadymagnetic field 0 B k� .Aparticle of mass �m�, positive charge �q� and velocity 0 v �i travels along x-axis and enters the region of themagnetic field. Neglect the gravitythroughout the question (a) Find the value of�L�iftheparticle emerges fromthe regionofmagnetic fieldwithits finalvelocityat anangle 30º to its initialvelocity. (b) Find the finalvelocity of the particle and the time spent by it in themagnetic field, if themagnetic field now expands upto 2.1 L. Sol. (A) . = 30º, sin LR . . Here 0 0 R mv qB . . 00 sin 30º L mv qB . ×+ + + + + + + + + + + + + + + + + + + + + + + P v0 X Y qACx=0 x=L v0 LR B = B0 k or 00 1 qB L 2 mv . . 00 L mv 2qB . (b) In part (i) sin 30º L

R . or 1 L 2 R . or L = R/2 × × × × × × × × × × × × B R v0 v0 L´ =2.1R > R Nowwhen L´ = 2.1 L or 2 2.1R 2 . L´ > R Therefore, deviation of the particle is . = 180º is as shown. . f 0 B v . .v i� . v . . and AB 0 t T m 2 qB . . . Example 27. Awire loop carrying a current �I�is placed in the x�y plane as shown in fig. (a) If a particlewith charge +Qandmass �m� is placed at the centre �P� and given a velocityv . alongNP (see figure), find its instantaneous acceleration. V P+Q 120º M I N a y O x (b) If an external unfiormmagnetic induction field B . B�i .. is applied find the force and the torque acting on the loop due to this field. Sol. Themagnetic field at the centre P due to current inwireNMis 0 . . 1 B µ I sin 60º sin 60º 4 r . . .

MAGNETIC FIELD www.physicsashok.in 18 0 1 µ B I 3 3 4 a / 2 2 2 . . . . . . . . . v P Q 120º MN a y 60º 30º 30º Q x V sin60º V0 V cos60º 1 B µ 2I 3 4 a . . directed awayfromthe reader perpendicular to the plane of paper. sin 30º ra . . r a2 . cos30º MS a . 30º S r P a MN a . MS 3 a 2 . . MN . 3 a Themagnetic field at the centre P due to current in arcMNis 0 0 0 2 µ 2 I µ 2 I 2 / 3 µ 2 I B 2 a 2 4 a 2 4 3a . . . . . . . . . . . . . . . . . . .. . . . . . directed towards the reader perpendicular to the plane of paper

The netmagnetic field 0 0 1 2 B B B µ 2 3I µ 2 I 4 a 4 3a . . . . . . . 0 0 µ 2I µ 2I B 3 (.68) 4 a 3 4 a . . . . . . . . . . . . . . (directed away fromthe reader perpendicular to the plane of paper) The force acting on the charged particleQwhen it has a velocity v and is instantaneously at the centre is F = QvB sin . = QvB sin 90º = QvB The acceleration produced 0 F QvB Qv µ 2I A (0.68) M m m 4 a . . . . . . . .. . .. 0 A 0.11µ IQv ma . y x P B v M R QQ N120º The direction of acceleration is given by the vecotr product v . B . .. or byapplying Fleming�s left hand rule

MAGNETIC FIELD www.physicsashok.in 19 ..RPN = 90º and .MPN = 120º . .MPR = 120º � 90º = 30º Since, .MPQ = 60º . .RPQ = 30º i.e.,wthe acceleration vectormakes an angle of 30º with the negative x-axis. The torque acting on the loop in themagnetic field is giving. . . M . B . ... .. where M= IA where A= (area of PMQNP)�(area of trangle PMN) A 1 . a2 . 1 MN PS 3 2 . . . . . 2 A a 1 3a a a2 3 3 2 2 3 4 . .. . . . . . . . . . . . A a2 3 k� 3 4 .. . . . . . . . .. . Ia2 3 k� �iB 3 4 .. . . . . . . . . . . BIa2 3 �j 0.614 BIa2J� 3 4 . . . . . . . . . . . . The force acting on the loop is zero. Example 28. An electron gunGemits electrons of energy 2 keVtravelling in the positiveX-direction. The electrons are requeired to hit the spot Swhere GS = 0.1m, and the lineGS make an angle of 60º with the x-axis as shown in figure.Auniformmagnetic field B.. parallel toGS exists in the region outside 60º B SG X the electron gun. Find theminimumvalue ofBneeded tomake the electrons hit S. Sol. Kinetic energyof electron, K 1 mv2 2keV 2 . . . speed of electron, v 2K m . 60º B S

G v 16 31 v 2 2 1.6 10 m/ s 9.1 10 . . . . . . . v = 2.65 × 107 m/s Since the velocity (v) . of the electronmakes an angle of . = 60º with themagnetic field B.. , the pathwillbe a helix. So, the particlewillhit S if GS = nP Here n = 1, 2, 3, ............. p pitch of helix 2 m v cos qB . . . .

MAGNETIC FIELD www.physicsashok.in 20 But for B to beminimum, n = 1 Hence, GS p 2 m vcos qB . . . . . min B B 2 mv cos q(GS) . . . . Substituting thevalues,we have 31 7 min 19 (2 )(9.1 10 )(2.65 10 ) 1B 2 tesla (1.6 10 )(0.1) . . . . . . . .. .. . . or Bmin = 4.73 × 10�3 tesla Example 29. An electronmovingwith a velocity 1 V . 2 i� .. m/s at a point in a magnetic field experiences a force 1 F . .2�jN . . If the electron ismoving with a velocity 2 V . 2�j .. m/s at the same point, it experiences a force 2 F . .2�i N . . The force the electronwould experience if it weremovingwith a velocity 3 V . 2 k� .. m/s at the same point is (A) zero (B) 2k.N (C) .2k.N (D) informationis insufficient Sol. F1 . qv1 . B . . ..1 .2 �j . .e(v ) . B . .. .2�j . .e(2 i�) . B.. �j . 2e( �i . B) .. �j . eB( �i) . (.k� ) . eB = 1 2 F . .e(2 �j) . (.Bk� ) . 2�i . 2eBi� eB = 1 F3 . .e(v3 . B) . . .. 3 F . .e(2 k� ) . (.Bk� ) . 0 . .

Hence option (A) is correct Charged particle in uniform & ... .. E B When a charged particlemoveswith velocityv . in an electric field E.. andmagnetic field B.. , then.Net force experienced byit is given byfollowing equation. F . qE . q(v . B) . .. . .. Combined forceF . is known as lorentz force (i) E || B||v .. .. . E B v In above situation particle passes underviated but its velocitywillchange due to electric field andmagnetic force on it is zero.

MAGNETIC FIELD www.physicsashok.in 21 (ii) E || B .. .. and uniform, particle is releasedwith velocity v0 at an angle .. v0 +q E, B v0 x E, B v0 cos +q v0 siny z 0 mv sin 2 m r ; T qB qB . . . . x . . 2 . . . .. . 0 r v cos t 1 qE t i� Rsin t �j R R cos �t k� 2 m . . . . . . . . . . . . . . . . R c y v0 sin z Cycloid motion Suppose that B.. points inthe x-direction, and E.. inthe z-direction. E x B z a b c y 0 F . q.E . v . B. . q.Ez� . Bzy� . Byz�. . ma . m.yy� . zz�. . .. . .. . . . .. .. qB m . . .. y z, z E y B . . . . .. . . .. . .. .. ... Their generalsolution is 1 2 3 2 1 4 y(t) C cos t C sin t (E / B)t C z(t) C cos t C sin t C . . . . . . .. . . . . . . y(t) E ( t sin t), B . . . . . z(t) E (1 cos t) B

. . .

. R EB . . (y � R.t)2 + (z � R)2 = R2 v R EB . . . The particlemoves as through it were a spot on the rimof awheel, rolling down the y axis at speed, v.The curve generated in thisway is called a cycloid. Notice that the overallmotion is not in the direction ofE.. , but perpendicular to it.

MAGNETIC FIELD www.physicsashok.in 22 Example 30. Aparticle of specific charge (charge/mass) . startsmoving fromthe origin under the action of an electric field 0 E . E �i .. andmagnetic field 0 B . B k� .. . Its velocity at (x0, y0, 0) is (4�i . 3�j) . The value of x0 is : (A) 0 0 13 E 2 B . (B) 0 0 16 B E. (C) 0 25 2.E (D) 0 5 2B. Sol. Theworkdone bymagnetic force isWB = 0, because,magnetic force is always perpendicular to instantaneous displacement. E 0 0 0 0 0 W . qE .S . qE �i . (x �i . y �j) . qE x .. . The speed of particle at (x0, y0) is v . 42 . (.3)2 . 5m/ s According to work � energytheorm,W T 1 mv2 1 mu2 2 2 . . . . or 2 E B W W 1 m5 0 2 . . . or 0 0 qE x 25m 2 . . 0 0 0 x 25m 25 2qE 2E . . . qm . . . . . .

. .

. Hence, option (C) is correct. Example 31. Aparticle of specific charge (q/m) is projected fromthe origin of coordinates with initial velocity [u �i . v �j] .Uniformelectricmagnetic fields exist inthe regionalong the+ydirection, ofmagnitude EandB. The particlewilldefinitelyreturn to theorigin once if (A) [vB/2.E] is an integer (B) (u2 + v2.1/2 [B/.E] is an integer (C) [vB/.E] in an integer (D) [uB/.E] is an integer Sol. y a qE m . mu2 quB r . r mu qB . also, T 2 m qB . . y vt qE t2 2m . . . For origin, x = 0, and y = 0 0 vt qE t2 2m . . .

MAGNETIC FIELD www.physicsashok.in 23 . t 2mV qE . For returning at the origin,nT 2mV qE . 2n m 2mV qB qE . . n 2mVqB qE(2 m) . . . n VBE . . Hence, option (C) is correct FORCE BETWEEN PARALLEL CURRENT CARRYING WIRES Consider two longwires 1 and 2 kept parallel to each other at a distance �r�and carrying currents i1 and i2 respectively in the same direction. The force per unit length of thewire 2 due towire 1 is : 0 1 2 F µ i i 2 r . l . F i1 l i2 1 2 The same force acts onwire 1 due to wire 2. r NOTE : The wires attract each other if currents in the wires are flowing in the same direction and they repel each other if the currents are in opposite directions. Example 32. Aconductor of length 2 mcarrying current of 2Ais held parallel to an infinitely long conductor carrying current of 10Aat a distance of 100mm. Find the force on small conductor. Sol. We knowthat force per unit length of short conductor due to long conductor is given by 0 1 2 f µ 2i i 4 r . . . Total force on length l of the short conductor is 0 1 2 F f µ 2i i 4 r . . . l l 7 10 2 2 10 2 5 F 8 10 N

0.1 . . . . . . . . . Force is attractive if the direction of current is same in both the parallel conductors and is repulsive if the direction of current is opposite intwo parallel conductors. Example 33. Astraight segment OC (of lengthL) of a circuit carrying a current �i�is placed along theX-axis as shown infigure.Two infinitely long straight wiresAand B, each extending fromz = � . to + ., are fixed at y = �a and y = + a respectively. as shown in the figure. If the ×× AB Y X O i C Z wiresAandB each carry a current �i�into the plane of the paper, obtain the expression for the force acting on the segment OC.What will be force ofOC if the current in the wire Bis reversed ?

MAGNETIC FIELD www.physicsashok.in 24 Sol. (a) Let us assume a segment of length dx at a point P, a distance x fromthe centre shown in figure. ×× AB y x O i y´ C I aa I dx x LBnet BB BA is the positive Z-direction and × is the negative Z-direction and Magnetic field at P due to current in wiresAand B will be in the directions perpendicular to AP and BP respectivelyas shown. 0 A B | B | | B | | B| µ i 2 AP . . . . .. .. .. Therefore, net magnetic field at Pwillbe along negative y= axis as shown and 0 net µ i x B 2 | B| cos 2 2 AP AP . . . . . . . .. ... .. .. .. 0 0 net 2 2 2 µ i.x µ i x B . (AP) (a x ) . . . . .. . .. . . Therefore, force on the element will be (F = ilB) 0 2 2 µ i x dF i dx a x . . . . . . . . . (in negative z-direction) . Total force onthewirewill be x L 2 L 0 2 2 x 0 0 F dF µ i xdx x a

.

.

. .

. . . . 2 2 2 0 2 µ i L a F ln 2 a . . . . . . . . . (innegative z-axis) Hence 2 2 2 0 2 µ i L a � F ln k 2 a . . . . . . . . . . . A× BY X i Bnet C BA BB P (b) If current inwire Bis reversed, thenmagnetic fields due to AandBwillbe in the directions shown in figure. i.e., net magnetic field net B .. willbe along positive x-axis and since current is also along positive x-axis, force onwireOCwill be zero. Note : A B .. is not necessarily parallel to BP. CURRENT LOOP IN UNIFORM MAGNETIC FIELD . . M . B . ... .. | . |. MBsin . . ; whereM= NIA M B I Work done in rotating loop in uniformfield from.1 to .2 W=MB (cos .1 � cos .2)

MAGNETIC FIELD www.physicsashok.in 25 Example 34. Themagneticmoment of a circular orbit of radius �r�carrying a charge �q�and rotatingwith velocity �v� is given by (A) qvr 2. (B) qvr 2 (C) qv.r (D) qv.r2 Sol. The convectioncurrent is I q qv 2 2 r . . . . . . Themagneticmoment is M I r2 qv r2 2 r . . . . . . M qvr 2 . Hence option (B) is correct Example 35. Qcharge is uniformly distributed over the same surface of a right circular cone of semi-vertical angle . and height �h�. The cone is uniformlyrotated about its axis at angular velocity.. Calculated associatedmagnetic dipolemoment. Sol. I = Themoment of gnertia3 MR2 10 . . Angularmomentum L I 3 mR2 10 . . . . . Butmagneticmoment P QL 2m . . P Q 3 mR2 2m10 . . . P 3Q R2 20 . . But tan Rh . . . R = h tan. . 3QR2 3Qh2 tan2 P 20 20 . . . . . Example 36. Figure shown a square current carrying loopABCDof side 10 cmand

current i= 10A. Themagneticmoment M... of the loop is (A) (0.05).�i . 3 k� .A.m2 (B) (0.05).�j. k� .A.m2 y x z 30º DC BA i = 10 (C) (0.05). 3 �i . k� .A.m2 (D) . �i . k� .A.m2 Sol. Themagnitude ofmagneticmoment is M = Il2 = 10 × (10 × 10�2)2Am2 = 10 × 10�2 = 0.1 Am2

MAGNETIC FIELD www.physicsashok.in 26 The normal on the loop is in x � z plane. It makes 60º angle with x - axis. . M . Mcos60º i� .Msin 60º �j ... . M M �i 3 M�j 2 2 . . ... . . . M 0.1 �i 3 �j 2 . . ... M . (0.05).i� . 3 �j.Am2 ... Example 37. Arectangular coil PQhas 2n turns, an area 2a and carries a current 2I, (refer figure). The plane of the coil is at 60º to a horizontaluniformmagnetic field of flux densityB. The torque onthe coil due tomagnetic force is 60ºB coil 2n, 2a, 2I (A) BnaI sin 60º (B) 8 BnaI cos 60º (C) 4 BnaI sin 60º (D) none Sol. . . M. B . ... .. Here M= 2n(2I) (2a) M= 8 nIa . . =MB sin(90º � 60º) . = MB cos 60º ..= 8 nIa cos 60º Hence option (B) is correct Example 38. (a)Arigid circular loop of radius �r�&mass �m�lies in the �xy�plane on a flat table and has a current �I�flowing in it.At this particular place, the earth�smagnetic field is x y B . B �i . B �j. .. .. .. Howlargemust �I�be before one edge of the loopwill lift fromtable ? (b) Repeat if, x y B . B �i . B k�. .. .. .. Sol. (a)Torque due to magnetic force should be greater than torque due to weight. . I.r2B . mgr or 2 2 2 x y I.r B . B . mgr .(b) Since, Bz is parallel to dipolemoment . 2 x I.r B . mgr . x I mg rB . . Example 39. Aconducting ring ofmass 2 kg and radius 0.5mis placed on a smooth horizontal plane.The ring carries a current i= 4A.Ahorizontalmagnetic field

B = 10T is switched on at time t = 0 as shown in figure. The initial angular B acceleration of the ringwillbe (A) 40. rad/s2 (B) 20. rad/s2 (C) 5. rad/s2 (D) 15. rad/s2 Sol. Aswe know, if a coilor a closed ofany shape is placed in uniform electric field,magnetic force on the coil or the loop is zero.

MAGNETIC FIELD www.physicsashok.in 27 So, the centre ofmass of the coil remains in rest. the torque on the coil is . . = IAB = 4.r2B or 2 mr 2 4 r B 2 . . . . 8B 8 10 40 rad / s2 m 2 . . . . . . . . Hence option (A) is correct Example 40. In the figure shown a coilof single turn iswound on a sphere of radius R andmass �m�. The plane of the coil is parallel to the plane and lies in the equatorial plane of the sphere. Current in the coil is �I�. The value of B if the sphere is in B equilibriumis (A) mg cos IR . . (B) mg .IR (C) mg tan IR . . (D) mg sin IR . . Sol. For equilibrium, mg sin . = f Also, net torque should be zero. Themagneticmoment of the loop is perpendicualr to the plane. . The torque due tomagnetic force is .B = PB sin(180 � .) 180º � P B . .B = PB sin. = I.R2B sin. This torque balances the torque due to friction about centre ofmass. . FR = I.R2Bsin. or mg sin . = .IRB sin . . B mg IR . . Hence option (B) is correct Example 41. Asquare current carrying loopmade of thinwire and having a massm= 10 g can rotatewithout frictionwith respect to the vertical axisOO1, passing throughthe centre of the loop at right angles to two opposite sides of the loop. The loop is placed in a homogeneousmagnetic

fieldwith an induction B = 10�1 T directed at right angles to the plane O+ B I O1 of the drawing.Acurrent I =2Ais flowing in the loop. Find the period of smalloscillations that the loop performs about its position of stable equilibrium. Sol. . = � PB sin. or . = � IAB. or I0. = � IAB. or 0 IAB I . . . . . . . . . . 2 0 IAB I . .. . . . or 0 IAB I . .

MAGNETIC FIELD www.physicsashok.in 28 or 0 2 IAB T I . . . T 2 I0 IAB . . Here 2 2 2 2 0 0 0 0 0 I m m m m 2 2 12 12 . . . . . . . . .. .. .. .. l l l l 2 2 0 0 0 m m I 2 6 . . l l But 3 0 m m 2.5 10 kg 4 . . . . l2 =A= area of loop, I = 2 amp After putting the value, T = 0.57 sec. Example 42.Auniformconstantmagnetic fieldB.. is directed at an angle of 45º to theX-axis inX�Yplane. PQRS is a rigid squarewire frame carrying a steady current I0, with its centre at the originO.At time t =0, the frame is at rest in the position shown in the figurewith its sides parallel toXandYaxes. Each side of the frame is of massMand length L. (a) What is the torque . . about Oacting on the frame due to the magnetic field ? (b) Find the angle bywhich the frame rotates under the action of this torque in a short intervalof time .t, and the axis about which this rotation occurs (.t is so short that any variation in Y 0 X P Q S R I the torque during this intervalmay be neglected).Given : the moment of inertia of the frame about an axis throughits centre perpendicular to its plane is 4ML2 3 . Sol. Magneticmoment of the loop, 2

0 M . (iA)k� . (I L )k� ... Magnetic Field, B (Bcos 45º )�i (Bsin 45º )�j B (�i �j) 2 . . . . .. (a)Torque acting on the loop, 2 0 M B (I L k� ) B (�i �j) 2 . . . . . . . . .. .. . ... .. . 2 0 I L B(�j �i) 2 . . . . or 2 0 | . |. I L B . (b)Axis of rotationcoincideswith the torque and since torque is in �j. �i directionor parallel toQS.Therefore, the loopwill rotate about an axis passing throughQand S as shown along-side : Angular acceleration, | | I. . . .

MAGNETIC FIELD www.physicsashok.in 29 Where I =moment of inertia of loop about QS. IQS + IPR = IZZ (Fromtheoremof perpendicular axis) Y X P Q S R But IQS = IPR . 2 QS ZZ 2I I 4ML 3 . . . 2 QS I 2ML 3 . . 2 0 0 2 | | I L B 3 I b I 2 ML 2 M 3 . . . . . . . Angle bywhich the frame rotates in time .t is 1 . ( t)2 2 . . . . or 0 2 3 I B . ( t) 4 M . . . MOVING COIL OR SUSPENDED COIL OR D´ ARSONVAL TYPE GALVANOMETER Principle : When a current-carrying coil is placed in magnetic field, it experiences a torque. Construction : It consists ofa narrowrectangular coilPQRS consisting of a large number of turns of fine insulated copperwirewound over a framemade of light, non-magneticmetal.Asoft ironcylinder known as the core is placed symmetricallywithin the coil and detached fromit. The coil is suspended between the two cylindrical pole-pieces (Nand S of a strong permanent horse-shoemagnet) by a thin flat phosphor bronze strip, the upper end ofwhichis connected to amovable torsion head T.The lower end of the coil is connected to a hair-spring �s� of phosphor bronze having onlya fewturns. N Core S P S Q R s T T1 T2 m Moving coil galvanometer In order to eliminate air-disturbance, thewhole arrangement is enclosed in a brass case having a glasswindow

on the front. Levelling screws are provided at the base.The torsion head T is connected to a binding terminal T1. So, the phosphor-bronze strip acts as one �current lead� to the coil. The lower end of the spring �s� is connected to a binding terminalT2.Aplanemirror or a concavemirror of larger radius of curvature is rigidly attached to the phosphor bronze strip. This helps to measure the deflection of the coil by lamp and scale arrangement. Radialmagnetic field : The magnetic field in the small air gap between the cylinderical pole-pieces is radial. Themagnetic lines of forcewithin the air gap are along the radii. On account of this, the plane of the coil remains N S always parallel to the direction ofthemagnetic field i.e., the angle between Radial magnetic field the plane of the coil and themagnetic field is zero in all the orientations of the coil.

MAGNETIC FIELD www.physicsashok.in 30 Theory : Let I = current flowing through the coil; B=magnetic field induction l = lengthof the coil; b = breadth of the coil N= number of turns in the coil; A(= l × b) = area of the coil Since the field is radial, therefore, the plane ofthe coilremains parallel to themagnetic field inall the orientations of the coil. So, the sides SP and QR remain parallel to the direction of the magnetic field. So, they do not experience any force.The sides PQandRS remain perpendicualr to the direction of themagnetic field. These sides experience forces perpendicular to the plane of the coil. b F F P S Q R l Current-carrying loop in magnetic field Force on PQ, F = NBIl Applying Fleming�s left hand rule,we find that this force is normal to the plane of the coiland directed outwards, i.e., towards the reader. Force on RS, F = NBIl Applying Fleming�s left hand rule,we find that this force is normal to the plane of the coiland directed inwards, i.e., awayfromthe reader. The forces onthe sides PQandRS are : (i) equal inmagnitude; (ii) opposite in direction; and (iii) act at different points. So, the two forces constitute a couple. This couple tends to deflect the coil and is known as deflecting couple. Moment of deflecting couple =NBIl × b =NBIA [The field is radial. The forces onthe sides PQandRS always remain perpendicular to the plane ofthe coil. So, the perpendicular distance between the forces is always equal to �b� as in fig.] b FP SF When the coildeflects, the suspension fibre gets twisted. On account ofelasticity, a restoring couple is set up in the fibre.This couple is proportional to the twist. If . be the angular twist, then Moment of restoring couple = k. where �k� is the restoring couple per unit angular twist. It is also known as torsional constant. For equilibriumof the coil, NBIA= k. or I k NBA . . . . .. .. or I . K. where K k NBA . . . .. .. is the galvanometer constant. Now, I .. or . . I

So, the deflection of the coilis proportional to the current flowing through the coil. This explains as towhywe can use a linear scale in a galvanometer.The scale is calibrated to give direct values of current.

MAGNETIC FIELD www.physicsashok.in 31 CURRENT SENSITIVITY OF A GALVANOMETER Agalvanometer is said to be sensitive if it gives a large deflection for a small current. The current sensitivity of a meter is the deflection of the meter per unit current, i.e., I . . It is given by NBA I k . . [. NBIA= k.] The sensitiveness can be increased by increasing N,Aand B and decreasing the value of �k�. But N andA cannot be increasedmuch because thiswill increase the length and consequentlythe resistance of the coil. In that case, the galvanometer will not respond toweak electric currents. B can be increased by using a strongmagnet. �k� can be decreased by using phosphor-bronze for suspension. �k� can be further reduced by using quartz suspension fibre. VOLTAGE SENSITIVITY OF A GALVANOMETER It is defined as the deflection of the meter per unit voltage, i.e., V. . Now, V RI . . . or NBA V kR . . Advantages : (i) The galvanometer can bemade extremely sensitive. (ii) Since themagnetic field B is very high, therefore, the externalmagnetic fields cannot appreciably after the deflection of the coil. So, the galvanometer can be used in any position. (iii) Since the deflection of the coil is proportional to current, therefore, linear scale canbe used. (iv) Since the coil is wound over metallic frame, therefore, damping is produced by eddy currents. So, the galvanometer coilcomes to rest quickly.This typeofgalvamometer is called aperiodic or dead beat galvanometer. The galvanometer can bemade ballistic bywinding the coil on a non-conducting frame of ivory or ebonite. (v) The lamp and scale arrangement used to measure the deflection of the coilmakes the galvanometer very sensitive. POINTER TYPE OR WESTON OR PIVOTED MOVING COIL GALVANOMETER The suspended typemoving coil galvanometers are very sensitive. Theycanmeasure currents of the order of 10�9 ampere. But these require very careful handling. So, for general use in the laboratory and for those experimentswhose sensitivityis not required, pointer type galvanometers are used. T1 T2 N S 30 20 10 0 10 20 30

MAGNETIC FIELD www.physicsashok.in 32 Inthis typeof galvanometer, the coilis pivoted between two ball-bearings.Alight aluminiumpointer is attached to themoving coil. The controlling couple is providedwith the help of a spring. NOTE : Numerical Examples based on Moving Coil Galvanometer Formulae used : 1. . = NBIA 2. k. = NBIA 3. Current sensitivity, a = NBA I k 4. Voltage sensitivity, a = NBA V kR Units used : B in tesla, A in m2, R in ohm, k in N m rad�1. EARTH�S MAGNETISM Afreelysuspendedmagnet always points inthe north-south direction evenin the absence ofanyothermagnet. This suggests that the earth itself behaves as amagnet which causes a freely suspendedmagnet (ormagnetic needle) to point always in a particular direction : northand south.The shape ofearth�smagnetic field resembles that of a barmagnet of length one-fifth of earth�s diameter buried at its centre. SN Magnetic axis Magnetic N-pole Geographic S-pole Magnetic equatorEquator Magnetic S-pole Geographic N-pole Geographic axis The south pole of earth�smagnet is towards earth�s north pole (eographical north), while the north pole of earth�smagnet is towards earth�s south pole (geographical south). Thus, there is a magnetic S-pole near the geographicalnorth, and amagneticN-pole near the geographical south.The positions of the earth�smagnetic poles are not well defined on the globe, they are spread over an area. Magnetic equator : The great circlewhose place is perpendicular to the earth�smagnetic axis is called earth�s magnetic equator. Geographical equator : The great circlewhose plane is perpendicular to geographical axis is called geographical equator. Magnetic meridian : The line joining the earth�smagnetic poles is called themagnetic axis and a vertical plane passing throughit is called themagneticmeridian. Geographicalmeridian : The line joining the geographical north and south poles is called the geographic axis and a vertical plane passing through it is called the geographicalmeridian. Magnetic Elements To have a complet knowledge of earth�smagnetismat a place, the following three elementsmust be known :

(i)Angle of declination (ii)Angle ofdip or inclination (iii)Horizontal component of earth�s field. (i) Angle of declination : The angle between the magnetic meridian and geographicalmeridian at a place is called the angle of declination (or simplythe declination) at that place.

MAGNETIC FIELD www.physicsashok.in 33 Magnetic meridian B H C D C´B´ Geographical meridian Geographic north Magnetic north In fig.ABCDis themagneticmeridian andAB´C´Dis the geographicalmeridian. The angleB´AB = . is the angle of declination. (ii) Angle of dip or inclination : The anglewhich the axis of needlemakeswith the horizontal, is called angle of dip (.). In otherwords, the angle ofdip at a place is the anglewhich the resultantmagnetic field ofearth at that placemakeswith the horizontal. N S C i V DA B H In fig.ACshows the direction of resultantmagnetic field of earth and the angleBAC(=..) between it and the horizontalABis the angle of dip. (iii) Horizontal component of earth�s field : The direction of earth�s field at themagnetic poles is normal to the earth�s surface (i.e., in vertical direction) and at magnetic equator it is parallel to the earth�s surface, (i.e., in horizontaldirection).Thus, the resultant earth�s field can be resolved in two components as shown in fig. (a) the horizontal component HalongABand (b) the vertical componentV, alongAD. Fromfig. Horizontalcomponent H = Be cos . ...(i) and vertical component V = Be sin . ...(ii) . ee V B sin tan H B cos. . . . . or V =Htan . AgainEqs. (i) and (ii) give 2 2 2 . 2 2 . e H . V . B cos . . sin . or 2 2 e B . H . V

MAGNETIC FIELD www.physicsashok.in 34 THINKING PROBLEMS 1. Of the three vectors in the equation F . qv.B, . . .. which pairs are always at right angles?Whichmayhave any angle between them? 2. If an electron is not deflected in passing through a certainspace, canwe be sure that there is no magnetic field in that region? 3. Abreamof protons is deflected side ways. Could this deflection be caused (a) by an electric field? (b) by a magnetic field ? (c) If either is possible, howcan you tellwhich one is present? 4. Arectangular current loop is in an arbitary orientation in an externalmagnetic field. Is anywork required to rotate the loop about an axis perpendicular to its plane? 5. Will a tangent galvanometer work in the polar region ? 6. At that deflections is a tangent galvanometermost sensitive? 7. Whyis themagnetic needle short in a tangent galvanometer ? 8. Why is the field in amoving coilgalvanometer radialin nature? 9. What is the greatest disadvantagewith a suspended-typemoving coilgalvanometer? 10. An ammeter is aWeston galvanometer whose resistance ismade negligible byshunting the coil. Is this true or false? 11. Avoltmeter is aWeston galvanometer ofveryhigh resistance. Is this true of False? 12. Which gives amore accurate value of a potential difference, a potentiometer or a voltmeter? 13. Why is an ammeter connected in series? 14. What is a faraday? 15. There is no charge in the energyof a charged particlemoving in amagnetic field althrough amagnetic force is acting on it. Is this true or false?Give reasons in support of your answer. 16. Acurrent-carrying circular conductor is placed in a uniformmagnetic fieldwith its plane perpendicular to the field.Does it experience anyforce? Ifit does,what is this force if its radius is a and the current passing through it equals? 17. Howdo you knowthat the current inside a conductor is constituted by electrons and not by protons? 18. Acopper pipe is filledwithan electrolyte.When a voltage is applied, the current in the electrolyte is constituted bythemovement of positive and negative ions in opposite directions.Willsuch a pipe experience a forcewhen placed in amagnetic field perpendicular to the current? 19. Two parallel wires currying current in the same direction attract each other while two beams of electrons travelling in the same direction repel each other.Explainwhy? 20. Can a charged particle entering a uniformmagnetic field normally fromoutside complete a circle? 21. Acylindrical electrolytic bath containing a solution of copper sulphate between two electrodes ismounted above the northpole of a strong electromagnet.One electrode is at the axis (�) and the other electrode is at the edge of the bath (+).What happens to the electrolyte in these circumstances? 22. Averystrong current ismade to flowfor a short time through a solenoid.Willthere be anychange in its length

and diameter?Explain. 23. Cosmic rays are charged particles that strike the atmosphere fromsome external source. It is found thatmore low-energycosmic rays eachthe earth at the northand southmagnetic poles thanat themagnetic equator.Why is this so ?

MAGNETIC FIELD www.physicsashok.in 35 SOLUTION OF THINKING PROBLEMS 1. The pairsF . , B.. and F . , v . are always at right angle. B.. and v . may have any angle between them. 2. No, we cannot be sure that there is no magnetic field because the force will be zero when the direction of motion is along the direction ofthe field. 3. (a)Yes, it could be due to an electric field directed perpendicular to the motion. (b)Ys, it could be due to a magnetic field. (c) Stop the proton and keep it stationary. Ifit sill experiences a force in the same direction, it is due to an electric field. If the force vanishes of stopping themotion, there is amagnetic field. 4. No, no work is done in rotating the coil becausework doneW=mB (1 �cos.). Here there is no charge in . and so nowork is done. 5. No, because the earth�smagnetic field is vertical there. 6. Atengent galvanometer ismost sensitivewhen the deflections are near about 0º. 7. The field due to the circular coil is uniformover a very small region about the centre of the coil. So the needle must be short so that it maybe assumed tomove in uniformmegnetic fields. 8. The field ismade radial in order to have a linear relation between the current and the deflection. 9. The greatest disadvntage is that this type of galvanometer is not portable. 10. It is true. 11. It is true. 12. Apotentiometer gives amore accurate value of a potential difference than a voltmeter as it draws no current fromthe cell. 13. Since the current following through the circuit has to pass through the ammeter, it must be connected in series because it is a characteristic property ofseries connection that the same current passes through allparts of the circuit. 14. Afaradayis the amount of charge required to liberate 1 gramequivalent of any sustance. 15. True.Acharged particle experiences a force at right angles to the velocity and so it moves in a circle with a constant speed. So there is no charge in the energyfo the particle though amagnetic force acts on it. 16. It does not experiencemayforce because the forces on the elements are directed radially away and sumup to zero. 17. BytheHall effect.When a current-carrying flet conductor is placed inamagnetic field perpendicular to the flat face, a transversevoltage is developed. Fromthe directionofthis voltage calledHallvoltage, it canbe determined whether the current is due to negative or positive charge carriers. 18. The current constituted bynegtive ions in the positive directionis the same as the current formed bythe positive ions and so both form current in the same direction; they do not cancel out each other. So the pipe will experience a force. 19. Acurrent inonwire produces onlyamagnetic field and on electric field (because a current-carrying conductor is electricallyneutral) over the other current-carrying conductor and so onlyama

gnetic forcewhich is attractive is nature arises between them. But an electron beamis a source of both, an electric and amagnetic field. So there arise bothmagnetic force (attractive) and electric force (repulsive) between them. The repulsive force being in excess of the attractive force, theyrepel each other. 20. No, the boundaryline of themagneitc field has to be the diameter of the circle,whatever be the velocity of the charged particle.Hence, it will complete a semi-circle. 21. It moves counterclockwise (viewed fromabove) 22. The current in the adjacent turns flow in the same direction and so they attract each other. Thus there is contraction along the length ofthe solenoid.As the same currents in the diametricallyopposite elements flowin opposite directions, they repel each other. On account of this repulsive force the diameter of the solenoid increases. 23. Because at the poles, themagnetic field is parallel to the direction ofmotion of the cosmic rayparticles, both being vertical. Hence the cosmic ray particles do not experience anyforce. The force experienced is given by F = qvb sin(v . ,B.. ).At the equator these particles experiencemaximumdeflecting force and hence low-energy particles cannot reach the earth.

MAGNETIC FIELD www.physicsashok.in 36 ASSERTION&REASON Astatement of Statement-1 is given and a Corresponding statement of Statement-2 is given just belowit of the statements,mark the correct answer as � (A) If both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation ofStatement-1. (B) If both Statement-1 and Statement-2 are true and Statement-2 isNOT correct explanation ofStatement-1. (C) If Statement-1 is true but Statement-2 is false. (D) If both Statement-1 and Statement-2 are false. (E) IfStatement-1 is false but Statement-2 is true. 1. Statement-1 : Magnetic field interacts with a moving charge and not with a stationary charge. Statement-2 : A moving charge produces a magnetic field. 2. Statement-1 : If an electron is not deflected while passing through a certain region of space, then only possibility is that there is no magnetic field in this region. Statement-2 : Force is directly proportional to magnetic field applied. 3. Statement-1 : Free electrons always keep on moving in a conductor even then no magnetic force act on them in magnetic field unless a current is passed through it. Statement-2 : The average velocity of free electron is zero. 4. Statement-1 : Electron cannot be accelerated by the cyclotron. Statement-2 : Cyclotron is suitable only for accelerating heavy particles. 5. Statement-1 : The coil is wound over the metallic frame in moving coil galvanometer. Statement-2 : The metallic frame help in making steady deflection without any oscillation. 6. Statement-1 : In electric circuits, wires carrying currents in opposite directions are often twisted together. Statement-2 : If the wire are not twisted together, the combination of the wires forms a current loop. The magnetic field generated by the loop might affect adjacent circuits or components. 7. Statement-1 : If an electron and proton enter in an electric field with equal energy, then path of electron is more curved than that of proton. Statement-2 : Electron has a tendency to form large curve due to small mass. 8. Statement-1 : If a proton and an . -particle enter a uniform magnetic field perpendicularly, with the same speed, the time period of revolution of . -particle is double that of proton. Statement-2 : In a magnetic field, the time period of revolution of a charged particle is directly proportional to the mass of the particle and is inversely proportional to charge of particle. 9. Statement-1 : If an electron while coming vertically from outer space enter the earth�s magnetic field, it is deflected towards west. Statement-2 : Electron has negative charge. 10. Statement-1 : An electron and proton enters a magnetic field with equal velocities, then, the force experienced by proton will be more than electron. Statement-2 : The mass of proton is 1837 times more than the mass of electron. 11. Statement-1 : The magnetic field produced by a current carrying solenoid is independent of its length and cross sectional area. Statement-2 : The magnetic field inside the solenoid is uniform.

12. Statement-1 : The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. Statement-2 : Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized. [JEE 2008]

MAGNETIC FIELD www.physicsashok.in 37 Level # 1 1. An electron is accelerated to a high speed down the axis of a cathode ray tube by the application of a potential difference of V volts between the cathode and the anode. The particle then passes through a uniform transverse magnetic field in which it experiences a force F. If the potential difference between the anode and the cathode is increased to 2 V, the electron will now experience a force (A) F 2 (B) F 2 (C) 2 F (D) 2 F 2. In a hydrogen atom, an electron of mass m and charge e is in an orbit of radius r making n revolutions per second. If the mass of the hydrogen nucleus is M, the magnetic moment associated with the orbital motion of the electron is (A) M .ner2 m (B) m .ner2M (C) (M m) ner2 m . . (D) .ner2 3. An electron of charge e moves in a circular orbit of radius r around a nucleus. The magnetic field due to orbital motion of the electron at the site of the nucleus if B. The angular velocity . of the electron is (A) 4 r2 0 eB . . . . (B) r0 eB . . . . (C) e4 rB 0 .. . . (D) e2 rB 0 .. . . 4. Three long, straight and parallel wires C, D and G carrying currents are arranged as shown in Figure. The force experienced by a 25 cm length D C G 30 A 10 A 20 A 3 cm 10 cm of wire C is (A) 0.4 N (B) 0.04 N (C) 4 x 10�3 N (D) 4 x 10�4 N 5. A charged particle of specific charge s passes through a region of space shown.

(A) Velocity of the particle in REGION 1 is B. . . . (B) Work done to move the charged particle in REGION 1 is ZERO. (C) The radius of the trajectory of the charged particle in REGION 2 is s..0 . . (D) The particle emerges from REGION 2 with a velocity ' . . v .1 . 2 B S E B0 xxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxx x x x x x x x x REGION 1 REGION 2 where ' . . = ... . 6. A wire of resistance R in the form of a semicircle lies on the top of a smooth table. A uniform magnetic field B is confined to the region as shown. The ends of the semicircle are attached to springs C and D whose other ends are fixed. E C DB B = 0 below this line . X If r is the radius of the semicircle and k is the force constant for each spring, then the extension x in each spring is (A) kR x . 2EBr (B) kr x . 2EBR (C) kR x . EBr (D) 2kR x . EBr 7. An infinite collection of current carrying conductors each carrying a current . outwards perpendicular to paper are placed at x = x0, x = 3x0, x = 5x0, ........ ad infinitum on the x axis. Another infinite collection of current carrying conductors each carrying a current . inwards perpendicular to paper are placed at x = 2x0, x = 4x0, x = 6x0, ....... ad infinitum Here x0 is a positive constant. The magnetic field at the origin due to the above collection of current carrying conductors is (A) ZERO (B) 4 x log 2 0 e 0 . . . (C) . (D) 0

0 e 2 x log 2 . . .

MAGNETIC FIELD www.physicsashok.in 38 8. A conductor ABCDEF, with each side of length L, is bent as shown. It is carrying a current . in a uniform magnetic induction (field) B, parallel to the positive y-direction. The force experienced by the wire is B A C D E L L Z B Y X O F (A) BIL in the positive y-direction. (B) BIL in the negative z-direction. (C) 3 BIL (D) zero 9. The square loop ABCD, carrying a current . , is placed in a uniform magnetic field B, as shown. The loop can rotate about the axis XX�. The plate of the loop makes an angle . ( . < 90°) with the direction of B. Through what A B B X X� Y Z CD . angle will the loop rotate by itself before the torque on it becomes zero ? . (A) . (B) 90° (C) 90° + . (D) 180° � . 10. An electron is fired from the point A with a velocity V0 = 2 x 108 m/s. The magnitude and direction of magnetic field that will cause the A B 15 cm x V0 y electron to follow a semicircular path from A to B is (A) 1.5 x 10�4 T out of the page. (B) 1.5 x 10�3 T into the page (C) 1.5 x 10�2 T into the page (D) 1.5 x 10�2 T out of the page 11. Two long straight parallel wires 2 m apart, are perpendicular to the plane of the paper. Wire A carries a current of 10 A directed into the plane of the paper. Wire B carries a current such that the magnetic field at P at a distance of 0.8 m from this wire is zero.The magnitude and direction of the current in wire is zero. The AB S 1.2 m

1.6 m 2 mP magnitude and direction of the current in wire B is (A) 2.8 A into the page (B) 2.8 A out of the page (C) 2.8 x 10�6 A into the page. (D) 2.08 A out of the page. 12. Two particles each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2 . . The rod is rotated at a constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is (A) 2mq (B) mq (C) m2q (D) mq . 13. If a charged particle is describing a circle of radius r in a magnetic field with a time period T then, (A) T 2 . r 3 (B) T 2 . r (C) T . r 2 (D) T . r 0 14. A non-conducting rod AB of length . has a linear charge density . . The rod is rotated about an axis passing through point A with constant angular velocity . as shown in the figure. The magnetic moment of A +++ B . the rod is (A) 2 2 ... (B) 2 3 ... (C) 3 3 2 .. . (D) 3 6 ...

MAGNETIC FIELD www.physicsashok.in 39 15. A charged particle of mass m and charge q is released from rest from (x0, 0) along an electric field 0 E j� . The angular momentum of the particle about origin (A) is zero (B) is constant (C) increases with time (D) decreases with time 16. Two particles Y and Z emitted by a radioactive source at P made tracks in a could chamber as illustrated in the figure. A magnetic field acted downward into the paper. Careful measurements showed that both tracks P Y were circular, the radius of Y track being half that of the Z track. Z Which one of the following statements is certainly true? (A) Both Y and Z particles carried a positive charge. (B) The mass of Z particle was one half that of the Y particle. (C) The mass of the Z particle was twice that of the Y particle. (D) The charge of the Z particle was twice that of the Y particle. 17. The resistances of three parts of a circular loop are as shown in the figure. The magnetic field at the centre O is (A) 0 6 I a . (B) 0 3 I a . (C) 0 23 I a . (D) Zero A BR C R a 2R O I 120° 120° 18. A particle with a specific charge s is fired with a speed v towards a wall at a distance d, perpendicular to the wall. What minimum magnetic field must exist in this region for the particle not to hit the wall? (A) v/sd (B) 2v/sd (C) v/2sd (D) v/4sd 19. Current . flows through a long conducting wire bent at right angle as shown in figure. The magnetic field at a point P on the right bisector of the angle XOY at distance r from O is (A) 0r .. .

(B) 0 2 r . . . X YO I P x 45° r (C) 0 . . 2 2 1 4 r .. . . (D) 0 . 2 1. 2 r .. . . 20. Two circular coils X and Y having equal number of turns and carry equal currents in the same sense and subtend same solid angle at point O. If the smaller coil X is midway d X Y d O be between O and Y, then if we represent the magnetic induction due to bigger coil Y at O as By and that due to smaller coil X at O as Bx, then (A) 1 yx BB . (B) y 2 x BB . (C) 12 yx BB . (D) 14 yx BB . 21. Two mutually perpendicular conductors carrying 1 . and 2 . lie in the x-y plane. Find locus of points at which magnetic induction is zero? (A) 22 2

y x . . . (B) 12 x y . . . (C) 12 y x . . . (D) 2 2 12 x y . . . .

MAGNETIC FIELD www.physicsashok.in 40 22. A coil in the shape of an equilateral triangle of side 0.02 m is suspended from a vertex such that it is hanging in a vertical plane between the pole pieces of a permanent magnet producing a horizontal magnetic field of 5 x 10�2 T. Find the couple acting on the coil when a current of 0.1 N S ampere is passed through it and the magnetic field is parallel to its plane. (A) 5 2N .m (B) 5 2N .10.4N .m (C) 5 3 .10.7N . m (D) 10.7N . m 23. A particle of charge q and mass m starts moving from the origin under the action of an electric field 0 E . E i� . and magnetic field 0 B . B k� . . It s velocity at (x, 3, 0) is .4i� . 3 �j. . The value of x is: (A) 0 0 36E B qm (B) 0 25 2 m q E (C) 0 10m q E (D) 0 0 25E B m 24. Two charged particles A and B enter a uniform magnetic field with velocities normal to the field. Their paths are shown in the Figure. The possible reasons are: (A) The momentum of A is greater than that of B (B) the charge of A is greater than that of B. (C) The specific charge of A is greater than that of B (D) the speed of A is less than that of B. 25. Two charged particle M and N enter a space of uniform magnetic field, with velocities, perpendicular to the magnetic field. the paths are as shown in the figure. The possible reason are: (A) the charge of M is greater than that of N. (B) the momentum of M is greater than that of N. (C) specific charge of M is greater than that of N. (D) the speed of M is less than that of N. 26. A current-carrying ring is placed in a magnetic field. The direction of the field is perpendicular to the plane of the ring, then: (A) there is no net force on the ring. (B) the ring may tend to expand. (C) the ring may tend to contact (D) none of these. 27. A charge q is moving with a velocity 1 v. .1i� m s at a point in a magnetic field and experiences a force . . 1 F . q .1�j .1 k� . N. If the charge is moving with a velocity 2 v. .1 �j ma/s at the same point, it experiences

MAGNETIC FIELD www.physicsashok.in 41 a force 2 F . q .1i� .1k�. N . . . . The magnetic induction B . at that point is: (A) .i� . �j . k�.Wb m2 (B) .i� . �j . k�.Wb m2 (C) ..i� . �j . k�.Wb m2 (D) .i� . �j . k�.Wb 28. A parallel beam of electrons is shot into a uniform electric field, initially parallel to and against the field with a small initial speed. Then: (A) the beam will pass through the field accelerating down the field without changing its width. (B) the beam tends to spread out at the beginning and to narrow down later. (C) the beam tends to narrow down at the beginning and to spread out later. (D) the total energy of the beam is conserved. 29. The ratio of the energy required to set up in a cube of side 10 cm a uniform magnetic field of 4 wb/m2 and a uniform electric field of 106 V/m is: (A) 1.4 x 107 (B) 1.4 x 105 (C) 1.4 x 106 (D) 1.4 x 103 30. A charged particle is fired at an angle . in a uniform magnetic field directed along the x-axis. During its motion along a helical path, the particle will: (A) never parallel to the x-axis. (B) move parallel to the x-axis once during every rotation for all values of . . (C) move parallel to the x-axis at least once during every rotation if . � 45°. (D) never move perpendicular to the x-direction. 31. A charged particle q enters a region of uniform B . (out of the page) and is deflected a distance d after travelling a horizontal distance a. The magnitude of the momentum of the particle is: (A) 2 2 qB a d d . . . . . . . (B) 2 qBa (C) Zero (D) not possible to be determined as it keeps changing. 32. Two insulated rings, one of slightly smaller diameter then the other, are suspended along their common diameter as shown in the figure, initially the planes of the rings are mutually perpendicular. When a steady current is set up in each of them: (A) The two rings rotate towards a common plane. (B) The inner ring will oscillate about its initial position. (C) The inner ring stays stationary while the outer one moves into the plane of the inner ring. (D) The outer ring stays stationary while the inner one moves into the plane of the outer ring. 33. A particle with charge +Q and mass m enters a magnetic field of magnitude B, existing only on the right of the boundary YZ. The direction of the motion of the particle is perpendicular to the direction of B. Let T 2 m

QB . . . The time spent by the particle in the field will be: (A) T. (B) 2T. (C) 2 2 T . . . . . . . . . . (D) 2 2 T . . . . . . . . . . 34. The current I flows through a square loop of a wire of side a. The magnetic induction at the centre of the loop is: (A) 0 2a . . . (B) 0 2 2a. . . (C) 0 2a. . (D) 0 2 a . . .

MAGNETIC FIELD www.physicsashok.in 42 35. Suppose a uniform electric field and a uniform magnetic field exist along mutually perpendicular directions in a gravity free space. If a charged particle is released from rest, at a point in the space: (A) particle can not remain in static equilibrium. (B) first particle will move along a curved path but after some time its velocity will become constant. (C) particle will come to rest at regular interval of time. (D) acceleration of the particle will never become equal to zero. 36. A circular loop of radius R carries a charge q uniformly distributed on it. It is rotated at a frequency f about one of the diameters. A uniform magnetic field B exists along its diameter . The maximum and minimum torques acting on the loop due to the magnetic field are, respectively: (A) . qfR2B, 0 (B) 0, 0 (C) 2. qfR2B , . qfR2B (D) None of these 37. A semi-circular wire of radius R is connected to a wire bent in the form of a sine curve to form a closed loop as shown n the figure. If the loop carries a current . and is placed in a uniform magnetic field B, then the total force acting on the sine curve is: B (A) 2B.R (downward) (B) 2B.R (upward) (C) B.R (upward) (D) Zero 38. A charged particle of unit mass and unit charge moves with velocity of v . .8i� . 6 �j.m s . in a magnetic field of B . 2 k�T . . Choose the correct alternative(s): (A) The path of the particle be x2 + y2 � 4x � 21 = 0. (B) The path of the particle may be x2 + y2 = 25. (C) The path of the particle may be y2 + z2 = 25. (D) The time period of the particle will be 3.14 s. 39. A proton, a deuteron and an .-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If rp, rd, and r. denote respectively the radii of the trajectories of these particles, then. (JEE 1997) (A) p d r r r . . . (B) d p r r r . . . (C) d p r r r . . . (D) p d r r r. . . 40. A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. The particle will move in a (JEE 1999) (A) straight line (B) circle (C) helix (D) cycloid 41. Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper, as shown. The variation of the magnetic field B along the line XX� is given by (JEE 2000) (A) d d X X� B (B) d d X X� B (C) d d X X�

B (D) d d X X� B 42. An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current . flows through PQR. The magnetic field due tot his current at the point M is H1. Now, another infinitely long straight conductor QS is connected at Q so that current is . 2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now H2. The ratio 1 2 H H is given by (JEE 2000) (A) ½ (B) 1 (C) 2 3 (D) 2

MAGNETIC FIELD www.physicsashok.in 43 43. An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along the +z direction, then (JEE 2000) (A) positive ions deflect towards +y direction and negative ions towards �y direction (B) all ions deflect towards +y direction. (C) all ions deflect towards �y direction (D) positive ions deflect towards �y direction and negative ions towards +y direction. 44. A non-planar loop of conducting wire carrying a current . is placed as shown in the figure. Each of the straight sections of the loop is of length 2a. The magnetic field due to this loop at the point x y z I 2a P(a, 0, a) points in the direction (JEE 2001) (A) 1 . � �. 2 . j . k (B) 1 . � � �. 3 . j . k . i (C) . . 1 � � � 3 i . j . k (D) 1 .� �. 2 i . k 45. Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to Athis plane. The speeds of the particles are V B A and VB respectively and the trajectories are as shown in the figure. Then (JEE 2001) (A) mA vA < mB vB (B) mA vA > mB vB (C) mA < mB and vA < vB (D) mA = mB and vA = vB 46. A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current . passes through the coil, the magnetic field at the center is (JEE 2001) (A) 0 N b . . (B) 0 2 N b . . (C) . . 0 ln 2 N b b a a . . . (D) . . 0 ln 2 N b b a a . .

. 47. A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters a region containing a uniform magnetic field B directed along the negative z direction, extending from x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b is (JEE 2002) (A) qbB m (B) q .b a. B m. (C) qaB m (D) . . 2 q b a B m . 48. A long straight wire along the z-axis carries a current . in the negative z direction. The magnetic vector field B . at a point having coordinates (x, y) in the z = 0 plane is (JEE 2002) (A) . . . . 0 2 2 � � 2 y i x j x y ... .. (B) . . . . 0 2 2 � � 2 x i y j x y ... .. (C) . . . . 0 2 2 � � 2 x j y i x y ..

. .

. (D)

. .

. . 0 2 2 � � 2 x i y j x y ... .. 49. For a positive charged particle moving in a x-y plane initially along the x-axis, there is a sudden change in its path due tot he presence of electric and/or magnetic fields beyond P. The curved path is shown x y in the x-y plane and is found to be non-circular. Which one of the P following combinations is possible? (JEE 2003) (A) E . 0; B . b i� . c k� . . (B) E . a i�; B . c k� . a i� . . (C) E . 0; B . c �j . b k� . . (D) E . a i�; B . c k� . b �j . .

MAGNETIC FIELD www.physicsashok.in 44 50. A conducting loop carrying a current . is placed in a uniform magnetic field pointing into the plane of the paper as shown. x Y I B The loop will have a tendency to (JEE 2003) (A) contract (B) expand (C) move towards +ve x-axis (D) move towards �ve x-axis. 51. A current carrying loop is placed in a uniform magnetic field in four different orientations, I, II, III & IV, arrange them in the decreasing order of Potential Energy (JEE 2003) B � n I B � n II B �n III B � n IV (A) I > III > II > IV (B) I > II > III > IV (C) I > IV > II > III (D) III > IV > I > II 52. An electron traveling with a speed u along the positive x-axis enters into a region of magnetic field where . . 0 B . .B k� x . 0 . It comes out B x ye u � of the region with speed v then (JEE 2004) (A) v = u at y > 0 (B) v = u at y < 0 (C) v > u at y > 0 (D) v > u at y < 0 53. A magnetic needle is kept in a nonuniform magnetic field. It experiences (JEE 1982) (A) a force and a torque (B) a force but not a torque (C) a torque but not a force (D) neither a force nor a torque 54. A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and magnetic fields respectively. This region of space may have:(JEE 1985) (A) E = 0, B = 0 (B) E = 0, B . 0 (C) E . 0, B = 0 (D) E . 0, B . 0 55. A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If steady current . is established in the wire as shown i I in the figure, the loop will : (JEE 1985) (A) rotate about an axis parallel to the wire (B) move away from the wire (C) move towards the wire (D) remain stationary 56. Two particle X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. The ratio of the mass of X to that of Y is (JEE 1988)

(A) . .1 2 1 2 R R (B) 2 1 R R (C) . .2 1 2 R R (D) 1 2 R R 57. A particle of charge +q and mass m moving under the influence of a uniform electric field E i� and uniform magnetic field B k� follows a trajectory from P to Q as shown in Figure. The velocities at P and Q are v i� and .2 �j 2v P a Q x 2a B V E which of the following statement(s) is/are correct? (JEE 1991) (A) 3 2 4 E mv qa . . . . . . . (B) Rate of work done by the electric field at P is 3 3 4 mv a . . . . . . (C) Rate of work done by the electric field at P is zero (D) Rate of work done by the electric field at Q is zero

MAGNETIC FIELD www.physicsashok.in 45 58. A current 1 flows along the length of an infinitely long, straight, thin-walled pipe. Then: (JEE 1993) (A) The magnetic field at all points inside the pipe is the same, but not zero (B) The magnetic field at any point inside the pipe is zero. (C) The magnetic field is zero only on the axis of the pipe (D) The magnetic field is different at different points inside the pipe. 59. H+ , He+ and O++ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity. The masses of H+, He+ and O2+ are 1 amu, 4 and 16 amu respectively. The: (JEE 1994) (A) H+ will be deflected most. (B) O2+ will be deflected most (C) He+ and O2+ will be deflected equally (D) All will be deflected equally. 60. Two very long, straight, parallel wires carry steady currents . and � . respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in the plane of the wires, Its instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is (JEE 1998) (A) 0 2 qv d . .. (B) 0 qv d ... (C) 0 2 qv d .. . (D) 0 61. An infinite current carrying wire passes through point O and in perpendicular to the plane containing a current carrying loop ABCD as shown in the figure. Choose the correct option (s) (JEE 2006) (A) Net force on the loop is zero C B O O´ A D (B) Net torque on the loop is zero. (C) As seen from O, the loop rotates clockwise. (D) As seen from O, the loop rotates anticlockwise. 62. A magnetic field 0 B . B �j .. exists in the region a < x < 2a and 0 B . .B �j ..

, in the region 2a < x < 3a, where B0 is a positive constant. A positive point charge moving with a velocity 0 v . v �j . , where v0 is a positive constant, enters the 0 a 2a 3a x �B0 B0 magnetic field at x = 0. The trajectory of the charge in this region can be like, (JEE 2007) (A) z a 2a 3a x (B) z x a 2a 3a (C) z a 2a 3a x (D) z x a 2a 3a 63. A particle of mass m and charge q, moving with velocity v enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field B perpendicular to the plane of the paper. The length of the Region II is l. Choose the correct choice(s) × × × × × × × × × × × × × × × × × × × × × × × × Region I Region II Region III 0 v (A) The particle enters Region III only if its velocity l v q B m . l (JEE 2008) (B) The particle enters Region III only if its velocity v q B m . l (C) Path length of the particle in Region II is maximum when velocity v q B m . l (D) Time spent in Region Ii is same for any velocity V as long as the particle returns to Region I

MAGNETIC FIELD www.physicsashok.in 46 FILL IN THE BLANKS 1. A neutron a proton, and an electron and alpha particle enter a region of constant magnetic field with equal velocities. The magnetic field is along the inward normal to the plane of the paper. The tracks of the X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X B C particles are labelled in Figure. The electron follows track ________ D and the alpha particle follows track ________. (JEE 1984) 2. A wire of length L meters carrying a current i amperes is bent in the form of a circle. The magnitude of its magnetic moment is ________ in MKS units. (JEE 1987) 3. In a hydrogen atom, the electron moves in an orbit of radius 0.5Å making 1016 revolutions per second. The magnetic moment associated with the orbital motion of the electron is ________ . (JEE 1988) 4. The wire loop PQRSP formed by joining two semicircular wires of radii R1 and R2 carries a current 1 as shown. The magnitude of II R2 S R C Q P R1 the magnetic induction at the centre C is ________ . (JEE 1988) 5. A wire ABCDEF (with each side of length L) bent as shown in figure and carrying a current . is placed in a uniform magnetic induction B B A C DE F Z X Y I parallel to the positively-direction. The force experienced by the wire is ________ in the ________ direction. (JEE 1990) 6. A metallic block carrying current . is subjected to a uniform magnetic induction as B . as shown in Figure. The moving charges experience a force F. given by _____ which results in the lowering of the potential of the face _____ Assume the sped of the carriers to be v. (JEE 1996) 7. A uniform magnetic field with a slit system as shown in figure is to be used as momentum filter for high-energy charged particles. With a field B Tesla, it is found that the filter transmits . -particles Source DetectorB each of energy 5.3 MeV. The magnetic field is increased to 2.3 B Tesla and deuterons are passed into the filter. the energy of each deuteron transmitted by the filter is _____ MeV. (JEE 1997)

MAGNETIC FIELD www.physicsashok.in 47 TRUE / FALSE 8. No net force acts on a rectangular coil carrying a steady current when suspended freely in a uniform magnetic field. (JEE 1981) 9. There is no change in the energy of a charged particle moving in magnetic field although a magnetic force is acting on it. (JEE 1983) 10. A charged particle enters a region of uniform magnetic field at an angle of 85° to the magnetic line of force. The path of the particle is a circle. (JEE 1983) 11. An electron and proton are moving with the same kinetic energy along the same direction. When they pass through a uniform magnetic field perpendicular to the direction of their motion, they describe circular paths of the same radius. (JEE 1985) TABLEMATCHING 12. Column I Column II (A) Electric field (P) Stationary charge (B) Magnetic field (Q) Moving charge (C) Electric force (R) Changes the kinetic energy (D) Magnetic force (S) Does not change kinetic energy 13. Regarding the trajectory of a charged particle, match the following: Column I Column II (A) In electric field (P) Straight line path (B) In magnetic field (Q) Circular path (C) In crossed field (R) Helical path (D) In mutually perpendicular electric and (S) Parabolic path magnetic field, charge being at rest (T) Parabolic path 14. A current flows along length of a long thin cylindrical shell: Column I Column II (A) Magnetic field at all points lying inside the shell (P) Inversely proportional with distance from axis of shell (B) Magnetic field at any point outside the shell (Q) Zero (C) Magnetic field is maximum (R) Just outside the shell (D) Magnetic field on the axis of the shell 15. Column I Column II (A) Unit of magnetic field (P) Am2 (B) Unit of magnetic permeability (µ0) (Q) N Am (C) Unit of magnetic flux (.) (R) N A2 (D) Unit of magnetic dipole moment (S) Nm A 16. Two long parallel wires carrying equal currents in opposite directions are placed at x = . a parallel to Y-axis with z = 0. Then: Column I Column II (A) Magnetic field B1 at origin O (P) 0 3 ia .. (B) Magnetic field B2 at P (2a, 0, 0) (Q) 0 4 ia ..

. (C) Magnetic field at M (a, 0, 0) (R) 0i a .. . (D) Magnetic field at N (�a, 0, 0) (S) Zero

MAGNETIC FIELD www.physicsashok.in 48 17. Two wires each carrying a steady current I are shown in four configurations in Column-I. Some of the resulting effects are deseribed in Column-II. Match the statements is Column-I with the statements in Column-II. Column I Column II [JEE 2007] (A) Point P is situated P (P) The magnetic fields (B) at P due to the currents midway between the wires. in the wires are in the same direction. (B) Point P is situated at the (Q) The magnetic fields (B) at P due to the currents in mid-point of the joining the the wires are in opposite directions. centers of the circular P wires, which have same radii. (C) Point P is situated at the (R) There is no magnetic field at P mid-point of the line joining the centres of the circular P wires, which have same radii. (D) Point P is situated at the (S) The wires repel each other. common center of the wires. P 18. Six point charges, each of the same magnitude q, are arranged in different manners as shon in Column-II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and µ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current. [JEE 2009] Column I Column II (A) E = 0 (P) Charges are at the corners of a regular hexagon. M is at the centre of the hexagon. PQ is + � + � � M + P Q perpendicular to the plane of the hexagon. (B) V . 0 (Q) � + � + � + Charges are on a line perpendicular to PQ at QPM equal intervals. M is the mid-point between the two innermost charges. (C) B = 0 (R) Charges are placed on two coplanar insulating rings at equal intervals. M is the common centre + � Q P M� � + + of the rings. PQ is rependicular to the plane of

the rings. (D) µ . 0 (S) Charges are placed at the corners of a rectangle of sides a and 2a and at the mid points of the P M Q +� + �� � longer sides. M is at the centre of the rectangle. PQ is parallel to the longer sides. (T) Charges are placed on two coplanar, insulating rings at equal intervals. M is the mid-point bet- P QM + + + � � � ween the centres of the rings. PQ is perpendicular to the line joining the centres and coplanar to the rings.

MAGNETIC FIELD www.physicsashok.in 49 PASSAGE TYPE PASSAGE = 1. A cyclotron is a device used by physicists to study the properties of subatomic particles. Small charged particles are deposited at high speed into a circular pipe. electric and magnetic field then accelerate the particle to an even higher speed. Finally, when the particle reaches the desired speed, a magnetic field keeps it moving in a circle at constant speed. In figure 1, amagnetic field pointing �into the page� keeps a charged particle travelling in a counterclockwise circle at constant speed inside the cyclotron. The magnetic force on the particle points towards the center of the circle, and has strength Fmag = qvB where q is the particle�s charge, v is its speed, and B is the magnetic field strength. As a result of this force, the particle moves in circles in the cyclotron at frequency 2 f qB. m . North West South East particle where m denotes the particle�s mass. The frequency (in hertz) is the number of revolutions completed by the particle per second. In the following questions, neglect gravity. 1. In figure 1, what is the direction (if any) of the particle�s acceleration? (A) North (B) East (C) West (D) It has no acceleration 2. If the magnetic field in figure were turned off, in which direction would the particle travel (until crashing)? (A) North (B) East (C) West (D) Now here; it would stop moving 3. An alpha particle has charge 2e and mass 4 amu. A proton has charge e and mass 1 amu. Let falpha and f proton denote the frequencies with which those particles circle a cyclotron. If both particles experience he same magnetic field in the same cyclotron, what is alpha proton f f ? (A) 2 (B) 1 (C) 1 2 (D) 1 4 4. In order for a cyclotron to work properly, the magnetic field must make the particle move in a circle. Which of the following particles would not work in a cyclotron? (A) lithium atom (Li) (B) positive lithium ion (Li+) (C) negative lithium ion (Li+) (D) all of the above particle would �work�. 5. Which of the following cannot create a magnetic field? (A) Electrical current flowing through a well-insulated straight metal wire. (B) A beam of electrons moving across a cathode ray tube. (C) Electric current flowing around a superconducting ring. (D) Static electricity (i.e., extra electrons) built up on a stationary door knob.

PASSAGE = 2. Magnetically levitated (MAGLEV) trains are considered to be important future travelling machines. The idea of MAGLEV transportation has been in existence since the early 1900s. The benefits of eliminating friction between the wheel and the rail to obtain higher speeds and lower maintenance costs has great appeal. The basic idea of a MAGLEV train is to levitate it with magnetic fields so that there is no physical contact between the train and the rails. For comparison, �bullet� trains in Japan have a maximum speed of about 250-300 km/h while a MAGLEV train under development has reached a speed of 411 km/hr. The MAGLEV train uses powerful on board superconducting electromagnets with zero electrical resistance to support the train above the rails.

(B) small size (D) zero electrical resistance. They have low maintenance cost but high running cost they have low maintenance cost but high energy efficiency they have low maintenance and running cost as well as low installation cost. (C) China (D) None of these (C) on the walls (D) on the station (B) small size (D) zero electrical resistance. They have low maintenance cost but high running cost they have low maintenance cost but high energy efficiency they have low maintenance and running cost as well as low installation cost. (C) China (D) None of these (C) on the walls (D) on the station another charged particle �2q of mass 2 m in a uniform magnetic field B as shown in figure. If the particles are projected towards each other with equal speeds v. 1. Find the maximum value of projection speed vm so that the two particles do not collide. 2. Find the time after which collision occurs between the particles if projection speed equals 2vm . 3. Assuming the collision to be perfectly inelastic find the radius of particle in subsequent motion. MAGNETIC FIELD The walls along the track contain a continuous series of vertical coils of ordinary wire. As the train passes each coil, the superconducting magnet on the train induces a current in these coils and makes them electromagnets. The electromagnets on the train and outside produce forces that levitate the train and keep it centered above the rails. In addition, electric current flowing through coils outside the train propels the train forward, as shown in Figure.

1. What is the major advantage of using super conductors in the electromagnets (A) very low resistance for electric current (C) low hysteresis loss 2. Choose the correct statement (A) MAGLEV trains have high maintenance cost (B) (C) (D) 3. (A) Japan 4. (A) MAGLE trains are operating in which country (B) USA Super conducting electromagnets are installed in (B) on the railroads Coils installed on the wall will carry currents only if train is moving very fast (B) all times whether train is standing or running only if train is moving irrespective of whether fast or slow. (D) None of these. PASSAGE = 3. A charged particle +q of mass m is placed at a distance d from xinside the train 5.

(A) (C) xxx

q,m -2q,2m VV xxxx

xxxx

d

xxxx

(Neglect the electric force between the charges) www.physicsashok.in 50

MAGNETIC FIELD www.physicsashok.in 51 Level # 2 1. A particle of charge q and mass m is projected from the origin with velocity v . = 0 . i�in a nonuniform magnetic field k�B . .B0x . . Here .0 and B0 are positive constants of proper dimensions. Find the maximum positive x coordinate of the particle during its motion. 2. A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is . . The disc rotates about an axis perpendicular to its plane passing through the centre with angular velocity .. Find the torque on the disc if it is placed in a uniform magnetic field B directed perpendicular to the rotation axis. 3. A wire loop ABCDE carrying a current . is placed in the x-y plane as shown in figure. A particle of mass m and charge q is projected from origin with velocity . .j�i� 2 V V . 0 . m/s. Find the (a) instantaneous acceleration. (B) If an external magnetic field i�B . B0 is applied, find the x yO A B CD E 90° force and torque acting on the loop due to this field. r/2 4. A long wire of radius a is placed along z-axis and carries current i as indicated in the figure. y-axis is taken perpendicular to the plane of paper directed into the paper. An electron escapes from the surface of the wire with velocity .0 directed along x-axis. Determine the maximum distance from the wire along x-axis up to X Z e O V0 i which electron can move. 5. A long horizontal wire AB, which is free to move in a vertical plane and carries a steady current of 20 A, is in equilibrium at a height of 0.01 m over another parallel long wire CD which is fixed in a horizontal plane and carries a steady A B current of 30 A, as shown in figure. Show that when AB is slightly depressed, C

D it executes simple harmonic motion. Find the period of oscillations. 6. A current . flows along a thin wire shaped as a regular polygon with n sides which can be inscribed into a circle of radius R. Find the magnetic induction at the centre of the polygon.Analyse the obtained expression at n.. . 7. Find the magnetic induction at the centre of rectangular wire frame whose diagonal is equal to d = 16 cm and the angle between the diagonals is equal to . = 30°; the current flowing in the frame equals . = 5.0 A. 8. A thin conducting strip of width h = 2.0 cm is tightly wound in the shape of a very long coil with crosssection radius R = 2.5 cm to make a single-layer straight solenoid. a direct current . = 5.0 A flows through the strip. Find the magnetic induction inside and outside the solenoid as a function of the distance r from its axis. 9. Find the magnetic moment of a thin round loop with current if the radius of the loop is equal to R = 100 mm and the magnetic induction at its centre is equal to B = 6.0 .T. 10. Calculate the magnetic moment of a thin wire with a current . = 0.8 A, wound tightly on half a tore (Figure). The diameter of the cross-section of the tore is equal to d = 5.0 cm, the number of turns is N = 500.

MAGNETIC FIELD www.physicsashok.in 52 11. A non-conducting thin disc of radius R charged uniformly over one side with surface density . rotates about its axis with an angular velocity . . Find : (a) the magnetic induction at the centre of the disc ; (B) the magnetic moment of the disc. 12. A small coil C with N = 200 turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in Figure. The cross-section area of the coil is S = 1.0 cm2, the length of the arm OA of the balance beam is . = 30 cm. When there is no current in the coil the balance is in equilibrium. On passing a current . = 22 mA through the coil the equilibrium is restored by putting the additional counterweight of mass .m = 60 mg on the balance pan. Find the magnetic induction at the spot where the coil is located. 13. A square frame carrying a current . = 0.90 A is located in the same plane as a long straight wire carrying a current .0 = 5.0 A. The frame side has a length a = 8.0 cm. The axis of the frame passing through the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which is . = 1.5 times greater than the side of the frame. Find : (a) Ampere force acting on the frame ; (B) the mechanical work to be performed in order to turn the frame through 180° about its axis, with the currents maintained constant. 14. In an electromagnetic pump designed for transferring molten metals a pipe section with metal is located in a uniform magnetic field of induction B (Figure). A current . is made to flow across this pipe section in the direction perpendicular both to the vector B and to the axis of the pipe. a B I Find the gauge pressure produced by the pump if B = 0.10 T, . = 100 A, and a = 2.0 cm. 15. A proton accelerated by a potential difference V = 500 kV flies through a uniform transverse magnetic field with induction B = 0.51 T. The field occupies a region of space d = 10 cm in thickness (Figure). Find the d . + B angle . through which the proton deviates from the initial direction of its motion. 16. A slightly divergent beam of non-relativistic charged particles accelerated by a potential difference V propagates from a point A along the axis of a straigth solenoid. The beam is brought into focus at a distance . from the point A at two successive values of magnetic induction B1 and B2. Find the specific charge q/m of teh particles. 17. A non-relativistic proton beam passes without deviation through the region of space where there are uniform transverse mutually perpendicular electric and magnetic fields with E = 120 kV/m and B = 50 mT. Then the beam strikes a grounded target. Find the force with which the beam acts onthe target if the beam current is equal to . = 0.80 mA. 18. A beam of non-relativistic charged particles moves without deviation through

the region of space A (Figure), where there are transverse mutually perpendicular electric and magnetic fields with strength E and induction B. When the magnetic field is switched off, the trace of the beam on the screen S A S.. shifts by .x . Knowing the distances a and b, find the specific charge q/m a b of the particles.

MAGNETIC FIELD www.physicsashok.in 53 Level # 3 1. An electron in the ground state of hydrogen atom is revolving in anticlock-wise direction in a circular orbit of radius R. (JEE 1996) (i) Obtain an expression for the orbital magnetic dipole moment of the electron. � n 30° B (ii) The atom is placed in a uniform magnetic induction B . such that the plane-normal of the electron-orbit makes an angle of 30° with the magnetic induction. find the torque experienced by the orbiting electron. 2. A current of 10 A flows around a closed path in a circuit which is in the horizontal plane as shown in the figure. The circuit consists of eight alternating arcs of radii r1 = 0.08 m and r1 = 0.12 m. Each arc subtends the same angle at the center. (A) Find the magnetic field produced by this circuit at the center. (JEE 2001) A D C r1 r2 (B) An infinitely long straight wire carrying a current of 10 A is passing the center of the above circuit vertically with the direction of the current being into the plane of the circuit. What is three force acting on the wire at the center due to the current in the circuit? What is the force acting on the arc AC and the straight segment CD due to the current at the center? 3. A wheel of radius R having charge Q, uniformly distributed on the rim of the wheel is free to rotate about a light horizontal rod. The rod is suspended by light inextensible strings and a magnetic field B is applied as shown in the figure. The initial tensions in the strings are T0. If the breaking tension of the strings are 3T0/2, find the maximum .0 B T0 T0 d angular velocity 0 . with which the wheel can be rotated. (JEE 2003) 4. A pair of stationary infinitely long bent wires are placed in the x-y plane as shown in Figure. The wires carry currents of . = 10 A each as shown. The segments LR and MS are along the x- axis. the segments RP and SQ are parallel to the y-axis such that OS = OR = 0.02 m. Find the magnitude and . . . . L R O P S M x y

Q direction of the magnetic induction at the origin O. (JEE 1989) 5. A solenoid of length 0.4 m and having 500 turns of wire carries a current of 3 A. A thin coil having 10 turns of wire and of radius 0.01 m carries a current of 0.4 A. Calculate the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid. (JEE 1990) 6. Three infinitely long thin wires, each carrying current in the same direction are in the x-y plane of a gravity free space. The central wire is along the y-axis while the other two are along x = . d. (JEE 1997) (a) Find the locus of points for which the magnetic field B is zero. (B) If the central wire is displaced along the z-direction by a small amount and released, show that it will execute simple harmonic motion. If the linear density of the wire is . , find the frequency of oscillations. 7. A proton and an alpha particle, after being accelerated through same potential difference, enter a uniform magnetic field the direction of which is perpendicular to their velocities. Find the ratio of radii of the circular paths of the two particles. (JEE 2004) 8. In a moving coil galvanometer, torque on the coil can be expressed as b = ki, whre �i� is current through the wire and �k� is constant. The rectangular coil of the galvanometer having numbers of turns N, area A and moment of inertia I is placed in magnetic field B. Find (a) �k� in terms of given parameters N, I, A and B. (b) the torsional constant of the spring, if a current i0 produces a deflection of ./2 in the coil in reaching equilibrium position. (c) the maximum angle through which coil is deflected, id charge Q is passed through the coil almost instantaneously. (Ignore the damping in mechanical oscillations) (JEE 2005)

MAGNETIC FIELD www.physicsashok.in 54 Answer Key ASSERTION&REASON Q. 1 2 3 4 5 6 7 8 9 Ans. A E A B A A D A B Q. 10 11 12 Ans. E B C Level # 1 Q. 1 2 3 4 5 6 7 8 9 10 Ans. C D C D ABCD C D B C C Q. 11 12 13 14 15 16 17 18 19 20 Ans. B A D D C A D A D C Q. 21 22 23 24 25 26 27 28 29 30 Ans. C C B BCD ACD ABC A BD C AD Q. 31 32 33 34 35 36 37 38 39 40 Ans. A A C B ACD B B ABD A A Q. 41 42 43 44 45 46 47 48 49 50 Ans. B C B D B C B A B B Q. 51 52 53 54 55 56 57 58 59 60 Ans. A B A ABD C C ABD B AC D Q. 61 62 63 Ans. AC A ACD FILL IN THE BLANKS / TRUE-FALSE / MATCH TABLE 1. D, B 2. 2 4iL. 3. 1.25 x 10�23 Am2 4. 0 1 2 1 1 4I R R . . . . . . . . 5. .lB ; +z direction 6. evB; ABCD 7. 14.0185 eV 8. True 9. True 10. False 11. False 12. [(A�PQ), (B�Q), (C�PQR), (D�QS)] 13. [(A�PT), (B�PQR), (C�P), (D�S)] 14. [(A�Q), (B�P), (C�R), (D�Q)] 15. [(A�Q), (B�R), (C�S), (D�P)] 16. [(A�R), (B�P), (C�Q), (D�Q)] 17. [(A�QR), (B�P), (C�QR), (D�QS)] 18. [(A�PRS), (B�RS), (C�PQT), (D�RS)] PASSAGE TYPES Passage = 1. 1. C 2. A 3. C 4. A 5. D Passage = 2. 1. D 2. C 3. D 4. A 5. C Passage = 3. 1. 2m V Bqd m . 2. Bq 6t m. . 3. R . 3d.

MAGNETIC FIELD www.physicsashok.in 55 Level # 2 1. B q 2mv 0 0 2. 4...BR4 3. (a) 0 . . . . 0 0 q � � a 4 2 B V j i 8 2 rm . . . . . . . (B) Zero, 0 2 B 21 4r ... .... . . 4. ei (2 mv ) ae 0 0 m .. . . 5. 0.2 Sec. 6. 0 nµ I tan 2R n. . 7. B 4 0 / d sin 0.10mT. . . . . . . 8. . . . . .. .. . . . . . . . . . . . . / 4 2 / r, r R. B /h (1 (h / 2 R) 0.3mT,r R, 0 2 0 9. 2 0 3 pm . 2.R B/. . 30mA.m 10. 2 2 pm . 1/ 2N.d . 0.5 A.m 11. (a) B 1/ 2 0 R . . .. (b) p 1/ 4 R4. m . ... 12. B . .mg. /N.S . 0.4 T. 13. (a) F 2 / .4 2 1. 0.40 N 0 0 . . .. . . . . . (b) A ( 0a 0 / ) ln [2 1) /(2 1)] 0.10 J. . . .. . . . . . . . 14. .p . .B/ a . 0.5 kPa. 15. 30 .

2mV arcsin dB q . . . . .. . . .. . . 16. 2 2 1 2 2 (B B ) q/m 8 V . . . . 17. F = mE . /qB = 20.N. 18. 2E xq/m a(a 2b)B2 . . . Level # 3 1. (i) M eh 4 m . . (ii) ehB 8 m . directed perpendicular to the both M ... and B.. . 2. (a) 6.54 x 10�5 Tesla (b) 0, Force on arc AC = 0, Force on segment CD = 8.1 × 10�6 N (inwards) 3. 0man 2 W DT BQr . 4. 1 x 10�4 T or Wb m�2. 5. . = 5.9 x 10�6 Nm. 6. (a) , z 0. 3 x . . d . (b) 1 2 0 d 2 . .. . . .. . .. . .. 7. p p p r m q 1 . r m q 2 . . . . .

8. (a) k = NAB, (b) 0 C . 2i NAB . ; (c) 0 Q NAB 21i . . �X�X�X�X�

ATOMIC PHYSICS

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

ATOMIC PHYSICS www.physicsashok.in 1 MASS OF A PHOTON Rest mass of photon is zero. Effectivemass : Ifwe assume thatmis the effectivemass of photons inmoving conditions, then according to Einstein�s theory. Energy of photon,E =mc2 m= 2 Ec but E = h. = pc, p =photon �momentum . m = 2 Ec = 2 hc. = pc C1: Drawthe graph of (a)momentumversus effectivemass of photon (b) Energyversus effectivemass, (c)wavelengthversusmomentum, (d) Effectivemass versuswavelength for a photon. Sol: . pO m . E O m (a) p = mc . straight line tan. = c (b) E = mc2 . straight line tan. = c2 . p . m (c)de-brogliewavelength of photon . = hp (d) Effectivemass of photon ..p = h . Rectangular hyperbola m = 2 Ec = 2 hc. = 2 hc .c m. = hc . Rectangular hyperbola Power of a light source Suppose P = Power of the light source, . = frequency of emitted photons, Energyof single photon= h. Let, n = numbers of photons emitted per sec by the source, then Thus, Energy emitted per sec. by the source =W= h. Intensity of light source at a point The amount ofenergyincident ona point per unit area at that point ina unit time is called the intensityoflight

at that point.

ATOMIC PHYSICS www.physicsashok.in 2 Thus, intensityat a point is energyincident per unit area unit time. Suppose, I = Intensityof light at a point . = Frequency of photon n0 = no. of photon incident / sec /Area . I = n0h. SI unit ofintensityis J/secondm2. Intensity of light at a distance r : Let, W= Power of a point source r Amount of energy received byspherical surface per second P = Amount of energy emitted by light source per second =W Hence, Intensityat distance r = W Area = 2 W 4.r Thus, I = 2 W 4.r Example 1: The intensityof sunlight on the surface of earth is 1400W/m2.Assuming themeanwavelength of sunlight to be 6000 Å, calculate the number of photons emitted fromthe sun per second assuming the average radius of earth�s orbit around sun is 1.49 × 1011m. Sol: Average radius, r = 1.49 × 1011m. intensity of sunlight received by earth = I = 1400W/m2, . =meanwavelength = 6000Å. Energy emitted per second by the sum= Power of the sun =W but, Power,W= nh.,where n is number of photons emitted bysun per sec. .....(i) Again, intensityat a distance �r�froma point source of powerW. I = 2 W 4.r Sun r Earth Power = P I = 2 nh 4 r. . [Fromequation (i)] n = I 4 r2 h . . . = I 4 r2 hc . . . [. h. = hc . ] . n = 4 r2 I hc . . =

2 22 7 34 8 4 (1.49) 10 6 10 1400 6.63 10 3 10. . .. . . . . . . . n = 1.18 × 1045 photons per sec. Photon Flux (i.e. photon/sec) Suppose,W= Power of a source. IfAis the area of themetal surface onwhichradiations are incident, thenthe power received bythe plate is W. = IA = 2 W 4 r . . .. . .. A If . is the frequency of radiation, then the energy is photon is given by h..

ATOMIC PHYSICS www.physicsashok.in 3 If n. is the number of photons incident on the plate per second, then, we have W. = n. h. 2 W 4 r . . .. . .. A=N. h. r P Area(A) . n. = Plate 2 W A 4 rh . . . . . . . . . . . . . n´ is called photon flux. Photon flow density (n0) : The number of photons incident on unit area of the plate in one second is called the photonflowdensity. i.e., Photon flowdensity, n0, = n ' A = 2 W 4.r h. Photon-concentration : The number ofphotons per unit volume of the space. If n is the number of photons incident per unit area per second, then Photon - concentration = nc , where c is the speed of light. Example 2: The power of light emitted bythe sun is 3.9× 1026W.Assuming themeanwavelengthof sunlight to be 6000 Å, calculate the photon flux arriving at 1m2 area on earth perpendicular to light radiations. The average radius of earth�s orbit around sun is 1.49 × 1011m. Sol: Power of light emitted by the sun,W= 3.9 × 1026W r = earth sunmean distance = 1.49 × 1011m . = meanwavelength of sun light = 6000 Å= 6 × 10�7m P Sun r Earth Intensityoflight an earth, I = 2 W 4.r Power received by areaAon earth,W. = I ×A . W. = 2 W 4.r × A but,W. = n. h., where n. is the number of photon incident per sec or photon-flux. . n. = W' h. = 2 W A 4 r h. . .

. . .

. . . . .

. .

. n. = 2 W A 4 r hc . . [..h. = hc/.] n´ = 26 7 2 22 34 8 (3.9 10 ) (6 10 ) 1 4 (1.49) 10 6.63 10 3 10 .. . . . . . . . . . . photon /sec = 4.22 × 1021 photon/sec

ATOMIC PHYSICS www.physicsashok.in 4 PRESSURE EXERTED BY PHOTONS OR RADIATIONS PRESSURE (a) Each photon carries energyandmomentum.Hencewhen photons of light is incident on a surface, the light is either absorbed or reflected or both is done by the surface. (b) change in momentumof light takes place, which causes impulse or force on the surface resulting in a pressure called radiation pressure. Let light of intensityI is incident on a surface. Each photon carries energy h. andmomentum= hc. . Energyincident onunit area in unit time = I(by the definition of intensity). Number of photons incident on unit area in unit time. i.e. N A.t = I h. , .....(i) Momentumdelivered to unit area inunit time, i.e., P A t .. = N A t . . . . . . . × (change inmomentumof each photon) .....(ii) Using equation (i)&(ii),we get P A t .. = I h . . . . . . . × (change inmomentumof each photon) but, P A t .. = Force exerted A , [. force exerted = Pt .. ] = Radiation pressure . Pr= Radiation pressure = I h. × (change inmomentumof each photon) .....(iii) Radiation pressure for perfectly absorbing surface : In this case, change ofmomentumsuffered by each photon= h

c. . Using the equation(iii),we get hv c Perfectly absorbing surface Radiation pressure = I h. × hc. = Ic . Radiation pressure = Ic for a perfectlyabsorbing surface. Radiation pressure for a perfectly reflecting surface : In this case, change inmomentumof each photon = hc. � hc. . . . . . . . = 2hc. hv c Perfectly reflecting surface hv Hence, using the equation (iii), we get c Radiation pressure = I h . . . . . . . × 2hc. . . . . . . = 2I c . Radiation pressure = 2I c for a perfectlyreflecting surface.

ATOMIC PHYSICS www.physicsashok.in 5 Radiation pressure for a surface of reflection coefficient (.) : hv c surface hv . c In this case,Momentumof incident photon= hc. Momentumof reflected photon= . hc. , . = reflection coefficient . Change inmomentum= hc. � hc. . . . .. . . . = (1 + .) hc. Thus, using equation(iii),we get, . Radiation pressure = (1 + .) Ic for photons falling normallyon a surface. NOTE : (i) For a perfectly absorbing surface . = 0 . Radiation pressure = Ic (ii) For a perfectly reflecting surface . = 1 . Radiation pressure = (1 + 1) Ic = 2I c C2: Alaser emits a light pulse of duration .= 0.13ms and energyE = 10J. Find themean pressure exerted by such a light pulsewhen it is focussed into a spot of diameter d = 10 .mon a surface perpendicular to the beamand possessing a reflection -coefficient . = 0.50. Sol: Laser energy, E = 10J Plate area = 4 Laser d d = 10 .m . d2 Pulse duration, . = 0.13 .s Pressure exerted bylight pulse i.e., Radiation pressure = (1 + .) Ic , by the definition of radiation pressure byphotons of light .....(i) Here, . = reflection � coefficient = 0.50 I = Intensityoflight c = speed of light = 3 × 108m/s

Bythe definitionof intensityof light, I = Laser energy (pulse duration).(Area of plate) i.e. I = 2 Ed 4. . . . . . . . .....(ii) . From(i)&(ii) Radiation pressure = (1 + .) 2 4E c. .d = (1 + 0.50) 8 3 12 4 10 3 10 (0.13 10. ) 100 10. . . . . . . = 4.9 × 107 Nm�2 = 490 atm

ATOMIC PHYSICS www.physicsashok.in 6 Example 3: 1 .Acurrent flows in an x-ray tube operated at 10,000 V. The target area is 10�4 m2. Find the pressure on the target, assuming that the electrons strike the target normallyand the photons leave the target normally.Consider the idealsituationwhere eachincident electron gives rise to a photon ofthe same energy. Sol: Energy of each electron = energy of each photon= E = 104 eV Momentumdelivered by each electron = p1 = 2mE (photon) p e� 1 p2 Plate Momentumtaken away byeach photon = p2 = Ec Change ofmomentumdue to each electron-photon pair = p1 � (�p2) = p1 + p2 Current incident on target = i= 10�6A . Number of electrons incident per second = i/e = no. of photons emitted per second. Totalmomentumchange per second = force = ie . . . . . . (p1 + p2) . Pr, Pressure = force / area = i Ae . . . . . . (p1 + p2) = i Ae . . . . . . 2mE Ec . . . . . . . . Pr = 6 4 19 10 10 1.6 10 . . . . . 4 19 31 4 19 1/ 2 8 (2 9.1 10 10 1.6 10 ) 10 1.6 10 3 10 . . . . . . . . . . . . . . . . . . = 3.7 × 10�6 Nm�2

FORCE EXERTED BY PHOTONS ON A SURFACE We know that,when photons of light is incident on a surface, the change inmomentumof photons takes place resulting in a force exertion by the photons on the surface. Let light ofpowerWis incident on a surface. Each photoncarries energy h. and,momentum= hc. .....(i) Energyincident in unit time =W(bythe definition of power) Power =W Surface But, power is given by W = N.t (h.), whereN= total no. of photons incident in time .t. N.t = Wh. .....(ii) Momentumdelivered inunit time. Pt .. = Nt . . . . . . . × (change inmomentumof each photon) = P h. (change inmomentumof each photon) but, Pt .. = rate ofchange ofmomentum

ATOMIC PHYSICS www.physicsashok.in 7 Hence, Pt .. = F = Force exerted by the photons. Thus, force = P h . . . . . . . × (change inmomentumof each photon) ...(iii) Force exerted by photons on a perfectly absorbing surface : In this case, change ofmomentumsuffered by each photon= hc. using equation(iii),we get hv c Perfectly absorbing surface force = Wh . . .. ... × hc. . . . . . . = Wc thus, F = Wc for a perfectly absorbing surface. Force exerted by photons on a perfectly reflecting surface: In this case, change ofmomentumsuffered by each photon = hc. � hc. . . . . . . . = 2hc. hv c surface hv . c using equation ( iii),we get Force = Wh . . .. ... × 2h

c. . . . . . . = 2Wc thus, F = 2P c , for a perfectlyreflecting surface. Force exerted by photons on a surface of reflection coefficient .: In this casemomentumof incident photon= hc. , hv c surface hv Momentumof reflected photon= . c . hc. . Change inmomentum = hc. � hc. . . . .. . . . = (1 + .) hc. Thus, using equation(iii),we get Force = h. . . . . . . . × (1 + .) hc. = (1 + .)Wc i.e. F = (1 + .)Wc for photons falling normally on a surface.

ATOMIC PHYSICS www.physicsashok.in 8 C3: If a point source of light of powerWis placed at the centre of curvature of a hemispherical surface,whose inner surface is completelyreflecting, then the force onthe hemisphere due to the light falling onit is given byF = W2c . W (Source) C4: If a perfectly reflecting solid sphere of radius r is kept in the path of a parallel beamof light of large aperture and having an intensityI, then the force exerted by the beamon the sphere is given F = pr2I c . I r Note that force is equal to the product of (I/c) and the projected area of the sphere. C5: If photon of light of powerWfalls at an angle . to a perfectly reflecting surface, then net force exerted on the plate is given by Fnet = 2Wc cos.. .. C6: Alaser emits a light pluse of durationT = 0.10ms and energyE = 10 J. Find themean pressure exerted by such a pulsewhen it is focused on a spot of diameter d =10 µmon a surface perpendicular to the beamand with a reflection coefficient . = 0.5. Sol. p,momentumof a photon = hv/c = E/c Momentumof reflected photon = .(E/c) . change ofmomentum= (E/c) � (�.E/c) = (1 + .)E/c Force exerted = [(1 + .) E/c]/T Pressure = 4 [(1 + .) E/c]/T.d2 = 2 8 4 10 4(1 )E 4(1 0.5) 10 cTd 3 10 10. 10. . . . . . . . . . . . = 6.37 × 106 Nm�2 C7: Aplane light wave of intensityI = 0.50Wcm�2 falls on a planemirror of reflection coefficient . = 0.8.The angle of incidence . = 45º. Find the normal presure exerted by light on that surface. Sol. If S is the area of the mirror onwhich light falls, the transverse section of the incident beamis S cos .. Momentumof the incident photons =(I/c)(S cos .).Normal component ofmomentum flux = (IS cos ./c) cos ..= IS cos 2./c. Momentumof reflected photons = .(IS cos2./c) in the opposite direction. . rate of change ofmomentum= force = (1 + .) IS cos2./c . normal pressure = force/area = (1 + .) I cos2./c . required pressure = (1 + 0.5) (0.5 × 104 cos2 45º)/3 × 108 = 1.25 × 10�5 Nm�2 Example 4:An isotropic point source of radiationpower Pis placed on the axis of an idealmirror.The distance between the source and themirror is n times the radius of themirro. Find the for

ce that light exerts on the mirror.

ATOMIC PHYSICS www.physicsashok.in 9 Sol. Energyfluxthrough the annular space between two cones ofhalf-angles . and . + d.= (P/4.) (2. sin . d.) . momentumof flux = P sin . d./2c d S nR BBRA . rate of change ofmomentumalong the normal. 2(P sin . d./2c) cos . = P sin . cos . d./c . force onmirror = 0 p sin cos d c . . . . . where . = half-angle of the cone subtened by themirror = tan�1(1/n) . F = P sin2 . / 2c = P/2(n2 + 1)c Example 5: Figure shows a small, plane strip ofmassmsuspended froma fixed support through a string.A continuous beamofmonochromatic light is incident horizontally on the strip and is completely absorbed. The energyfalling on the strip per unit time isW. find the deflection of the string fromthe vertical, ifthe strip stays inequilibrium. Sol: Force exerted by the photons of light = (number of incident photons per sec) × (change inmomentumof each photon) Let, Number of incident photons per sec =N Light change inmomentumof each photon= hc. hence force exerted by the photons of light, F = N hc. = (Nh ) c . but, W = Nhv, by the definition of power . F = Wc If the stringmakes an angle .withthe verticalwhen the strip is at equilibrium, thenfor the equilibriumofthe strip, Tsin. = F, in horizontaldirection, . T F m mg T = tension in the string and Tcos. =mg, in verticaldirection. . Dividing the above equations,we get

tan. = F mg = W/c mg = W cmg , . tan. = W cmg .... = tan�1 W cmg . . .. .. Example 6. Aplane light wave of intensity I = 0.80Wcm�2 illuminates a sphere of radiusR = 5.0 cm. Find the force that the light exerts on the sphere. Sol. p,momentumof the incident pulse = E/c. i p . (E / c) (sin .�i . cos.�j) . with respect to a frame of reference with the outward normal as the y-axis and a line perpendicular to it and lying in the plane of themirror as the x-axis.

ATOMIC PHYSICS www.physicsashok.in 10 pf = .E/c and f p . (.E/ c) (sin. �i . cos.�j) . f i . p . p . p . (..1)E / c sin .�i . (..1) cos.E / c �j . . . . | . p | . E / c (. .1)2 sin2 . . (..1)2 cos2 . . . E / c 1. .2 . 2.cos 2. | . p | . .10 / 3 .108 . 1. 0.82 . 2 . 0.8cos60º . = 5.2 × 10�8 Nm�2 Impulse applied by photon on a surface Let hn be the energy of photons of a light incident normallyon a surface, Momentumof anincident photon= hc. . Change inmomentumof the photon takes place due to impact ofthe photonswith the surface. This change inmomentumofthe photons causes impulse. Using impulse �momentumtheorem,we get Impulse = Total change inmomentumof photons = (Total number of photons) × (change inmomentumof each photon) . Impulse =N. × (change inmomentumof each photon) ......(i) Impulse on a totally absorbing surface: In this case, change inmomentumof each photon = hc. . using equation(i),we get Impulse = N'h c . = Ec , where E is the total energyof the light . Impulse = Ec for a perfectlyabsorbing surface. Impulse on a totally reflecting surface: Impulse = N. × (change inmomentumofeach photon) = N. × h h c c . . . . .. . . .. .. . . .. = 2(N'h ) c . = 2E c [. E = total energy=N.h.] . Impulse =

2E c for a perfectlyreflecting surface. Impulse on a surface of reflection-coefficient .: In this case, change inmomentumof each photon = hc. � hc. . . . .. . . . = (1 + .) hc. . Using equation(i),we get

ATOMIC PHYSICS www.physicsashok.in 11 Impulse = N.(1 + .) hc. = (1 + .) N'h c . . . . . . . = (1 + .) Ec , [E =total energy] . Impulse =(1 + .) Ec for photons falling normally on the surface. Example 7: A small perfectly reflecting mirror of mass m = 10 mg is suspended byaweightless thread oflength . = 10 cmas shown in the figure. Find the angle throughwhichthe threadwillbe deflectedwhen a short laserwith energyE =13J is shot in horizontaldirection at right angles to themirror. (g = 10m/s2). m m Laser . Sol: Impulse applies bylaser on themirror, Impulse= 2E c , asmirror is perfectly reflecting. Initialmomentumof themirror = 0 v = 0 v . . .(1 � cos.) = H Let, finalmomentumof themirror =mv change inmomentumof themirror due to impact =0 � (�mv) =mv,where v is the speed ofmirror just after impact. But, Impulsemomentumtheoremgives, Impulse =change inmomentum 2E c = mv or v = 2E mc .....(i) Totalmechanical energyof themirror willbe conserved after impact. Thus, Loss in kinetic energy=Gainin potential energy 12 mv2 = mgh ..... v2 = 2gH 2 2 2 4E m c = 2gH, [as v = 2E mc fromequation (i)] 2 2 2 2E m c

= gH . 2 2 2 2E m c = g.(1 � cos.), [. H = . (1 � cos.)] 1 � cos. = 2 2 2 2E m c g. 2sin2 2. . . . . . . = 2 2 2 2E m c g. [. 1 � cos. = 2sin2 2. . . . . . . ] sin 2. = E2 mc g. Here E = 13J, m= 10 × 10�6 kg, . = 0.1 m . sin 2. = 5 8 13 10. .3.10 10.0.1 = 0.0043 or . = 0.5º PHOTONS UNDER GRAVITY

ATOMIC PHYSICS www.physicsashok.in 12 Photons can be considered as a particle ofmassm. If m = mass of photon v = frequencyof photon E = energy ofphoton, then mc2 = E = hv . m = 2 Ec = 2 hc. Thus, a photonof frequency v acts gravitationally like a particle ofmass 2 hc. . C8: If a photon frequency v falls on the surface of earth froma height h, thenwhat will be its frequency on the surface of earth. Sol: Change in frequencyof the photontakes place. v, E = hv + mgH v,. E = hv Ground H Let v. be the frequency of photon on the surface of earth. Mass of photon = 2 hc. Mass of the photon depends onits frequency, but wewill consider themass to be constant as difference in v´ and v is very small. Energyconservation gives Initial energy= final energy . hv + mgH = hv. . hv + 2 hc. gH = hv. . v. = v 2 1 gH c . . . . . . . Example 8:Aplanet ofmassMand radius R emits a photon of frequency v.What will be the frequency of photonwhen it covers an infinite distance. Sol: Let v..be the frequencyphoton at a infinite distance. v. R at . M G=Gravitational constant v Mass of the photon, m= 2 hc. Energy of photon on the surface = hv �G Mm

R = hv � 2 GM h R c. .. .. .. Energyof photonat infinity= hv. Energymust be conserved, . hv � 2 GM h R c. .. .. .. = hv´ . v. = v 2 1 GM Rc . . . .. .. NOTE: .. . = 2 GM Rc is called frequency shift. PHOTOELECTRIC EFFECT

ATOMIC PHYSICS www.physicsashok.in 13 Ejectionofelectrons fromametalplatewhenilluminated bylight or anyother radiationofsuitablewavelength (or frequency) is called photoelectric effect. This phenomenonwas first discovered byHeinrichHertz in 1887.One year later,Hallwachsmade the important observation thatwhen a negatively-charged zinc plate is irradiatedwith ultra-violet light, it loses its negative charge.Afterwards, itwas discoveredthat alkalimetals like lithium, sodium, potassium, rubidium and caesiumeject electronswhen visible light falls on them.Ten years later, J.J. Thomson and P.Lenard showed that the action of light was to cause the emission of free electrons frommetal surface.Although these electrons are no different fromallother electrons, it is customaryto refer to themas photoelectrons. Experimental Study of Photoelectric Effect Quartz C A � + E V Photoelectric phenomenoncanbe studiedwith the help ofa simple apparatus showninfigure. It consists oftwo photosensitive surfaces E andC enclosed in an evacuated quartz bulb. In the absence of light, there is no flowof current in the circuit and the ammeterA reads zero. When plate E is exposed to some source of monochromatic light, current starts flowing.However, no current is found to flowwhen light falls on plate C. The explanation of the above behaviour lies in the fact that whenE is irradiated with light, the incident photons eject electrons by collisionwith its atoms. These photoelectrons are immediatelyattracted bythe collector plateC thereby starting current flowas indicated bythe ammeter. WhenC is irradiated, even then photoelectrons are produced but theyare not allowed to leave plate C (i) firstly, because of the pulling effect of positive potentialofCand (ii) secondly, due to repulsion fromthe negative plate E. Hence, no current is found to flowin the circuit. Einstein�s Photoelectric Equation: Following Planck�s idea that light consists of photons, Einstein proposed an explanation of photoelectric effect as early as 1905.According to this explanationwhena single photon is incident on ametalsurface, it is completely absorbed an imparts its energyhf to a single electron. The photon energy is utilised for two purposes: (i) Partlyfor getting the electronfree fromthe atomand awayfromthemetal surface. This energyis known as the photoelectricwork function of themetal and is represented byW0. (ii) the balance of the photon energyis used up in giving the electron a kinetic energyof 1/2mv2. . hf =W0 + 1/2mv2 .....(i) It is knownasEinstein�s photoelectric equation. In case, the photon energy is just sufficient to liberate the electrononly then no energywould be available for imparting kinetic energy to the electron.Hence, the above equationwould redu

ce to hf0 = W0 .....(ii) where f0 is calledthe thresholdfrequency. It is definedas theminimumfrequencywhichcancause photoelectric emission. For frequencies lower than f0, therewould be no emission of electronswhereas for frequencies greater than f0, electronswould be ejectedwith a certain definite velocity(and hence kinetic energy). Substituting this value ofW0 in equation (i) above, the Einstein�s photoelectric equation becomes hf = hf0 + 12 mv2 or hf = hf0 + K.E. Long Wavelength Limit (.0)

ATOMIC PHYSICS www.physicsashok.in 14 It is thewavelength corresponding to the threshold frequency f0. Its physical significance is that radiations havingwavelength longer than .0would not be able to eject electrons froma givenmaterialwhereas those having . < .0will. Inotherwords, it represents the upper limit ofwavelengthfor photoelectric phenomenon. By analogy, it is also referred to as thresholdwavelength. Now, c = f0 .0 ...0 = 0 c f Also, W0 = hf0 . 0 1f = 0 h W . .0 = 0 ch W (i) WhenW0 is in joules .0 = 8 34 0 3 10 6.625 10 W . . . . = 26 0 19.875 10 W . . metre (ii)WhenW0 is electron volts (eV) .0 = 8 34 19 0 3 10 6.625 10 1.602 10 W . . . . . . = 7 0 12.4 10 W . . metre = 0 12, 400

W Å Kinetic energy of Photoelectrons Einstein�s photoelectric equationcan be used to find the velocityand hence the kinetic energyof an ejected photo-electron. hf = W0 + 12 mv2 = hf0 + K.E. . K.E. = hf � hf0 = h(f � f0) Now, f = c. and f0 = 0 c . . K.E. = ch 0 . 1 1 . . . . . . . . = 3 × 108 × 6.625 × 10�34 0 . 1 1 . . . . . . . . = 19.875 × 10�26 0 . 1 1 . . . . . . . . joules . and .0 inmetres = 19.875 × 10�26 0 . 1 1 . . . . . . . . joules . and .0 in Å

ATOMIC PHYSICS www.physicsashok.in 15 = 16 19 19.875 10 1.602 10 . . . . 0 . 1 1 . . . . . . . . eV . and .0 in Å . K.E. = 12,400 0 . 1 1 . . . . . . . . eV . and .0 in Å Incidentally, itmaybe noted that this also represents themaximumvalue ofthe kinetic energya photoelectron can have. . Emass = h(f � f0) = h..f joules = 12,400 0 . 1 1 . . . . . . . . eV . and .0 in Å Photoelectric Work Function As explained above it is defined as the energywhich is just sufficient to liberate electrons froma bodywith zero velocity. Its value is given by W0 = hf0 = 0 ch . = 8 34 0 3.10 .6.625.10. . = 26 0 19.875.10. . joules .0 inmetres = 16 0 19.875.10. . joules .0 in Å = 16 19 0 19.875 10 1.602 10 . . .

. . eV .0 in Å

. W0 = 0 12, 400 . eV .0 in Å Laws of Photoelectric Emission R B V Quartz C A The apparatus shown in figurewas used byMillikan to studythe photoelectric effect and the various laws governing it. S is a source of radiations of variable but known frequency f and intensity I. E is the emitting electrodemade of the photosensitivematerialand C is the collecting electrode. Both these electrodes are enclosed in an evacuated glass envelopewith a quartzwindowthat permits the passage of ultraviolet and

ATOMIC PHYSICS www.physicsashok.in 16 visible light.As shown, anypotentialdifference can be established between the two electrodes.Areversing switch RS enables the polarities of the two electrodes to be reversed. If the tube is in the dark, then no photoelectrons are emitted and themicroammeterAread zero. However, if ultraviolet or visible light is allowed to fall on the emitting electrode, electrons are liberated and circuit current is set up. Fromthe experimentaldata collected byRichardson and Compton in 1912, photoelectric emissionwas found to be governed bythe following laws: (i) Photoelectric current (i.e., number of electrons emitted per second) is directly proportional to the intensityofthe incident light (or radiation). Light Intensity I Photoelectric current Frequency constant This can be verified by increasing the intensity of light and measuring the corresponding photoelectric current while holding the frequency of the incident light frequency of the incident light and the accelerating potential V of the collecting electrode C constant. The graph is similar to one shown in figure. Increase in intensity means more photons and hence greater ejection of electrons. (ii) For each photosensitive surface, there is aminimumfrequency of radiation (called threshold frequency) at whichemission begins. f0 Frequency Photoelectric current Intensity constant Light oflower frequency(or longerwavelength), however strong, willnot be able to produce any electron emission. This fact can be verified by keeping the light intensity constant while varying the frequency.The graph so obtained is shownin figure.The current is found to increase (though non-linearly) with the frequency of the incident light. Moreover, it is seen that there is a limiting or critical frequency below which no photoelectons are emitted. It is called threshold frequency and its value depends on the nature of thematerial irradiated because for eachmaterial there is a certain minimumenergynessecaryto liberate anelectron. This energy is known as photoelectricwork function or threshold energyW0.As seen fromf0 =W0/h. Thewavelengthcorrespondingto be thresholdfrequencyf0 is called thresholdwavelengthor longwavelength limit.No photo-electrons are emitted for wavelength greater than .0, nomatter howlong the light falls on the surface or howgreater is its intensity.The photoelectric or quantumyield (which is definedas the photoelectric current in amperes per watt of incident light) depends on the frequency (and not intensity) of the incident light. (iii) Themaximumvelocityofelectron emission (and hence kinetic energy) varies

linearlywith the frequencyo the incident light but is independent of its intensity. As seen fromEinstein�s photoelectric equation of 2max 1 mv 2 = h(f � f0) or Emax . f f0 Frequency Emax . Intensity constant Hence, increase in the frequency ofthe incident light increase the velocitywithwhich photoelectrons are ejected.The same fact is illustrated byfigure. Incidentally, itmay be noted that the slope of the curve gives the value ofPlanck�s constant h. If the intensityof the incident light is increased (keeping frequencyconstant),more photonswillbe incident

ATOMIC PHYSICS www.physicsashok.in 17 on themetal surface, each photon having the same energy.Hence,more photo electronswill be ejected. Since an electron can absorb only one photon, each photoelectronwillhave the same energyandwill be ejectedwiththe same velocity.Obviously, increase inintensityonlyincreases the number of photo-electrons ejected but not their kinetic energy. (iv) The velocities of emitted electrons have values betweenzero and a definitemaximum. The proportion of the electrons having a particular velocityis independent of the intensityof radiation. (v) Electrons are emitted almost instantaneously evenwhen the intensityof incident radiations is very low. The time lag between the incidence of radiation and emission of first electrons is less than 10�8 second. (vi) For a givenmetalsurface, stopping potentialV0 isdirectlyproportional to frequencybut is independent of the intensityof the incident light. v i v0 O a b IH Suppose in figure, the frequencyand intensityofincident light are held constant but the potential difference V between the two electrodes E and C is increased. Up to some stage as this p.d. is increased photoelectric current is also increased.However, soon some value ofVis reachedwhen the current reaches a limiting or saturation value when al the photoelectrons emitted by E are immediately collected byC. Further increase inVhardlyproduces anyappreciable increase in current as shown by the flat portion of curve I in figure. IfVis reversedwith the help of the reversing switchRS i.e. E ismade positive and Cnegative, the current i does not immediatelydrop to zero proving that electrons are emitted fromE with some definite velocity. This velocity is such that it gives enough kinetic energy to the electrons so as to surmount the retarding electric field between the two electrodes.Hence, some electrons do succeed in reachingCdespite the fact that the electric field opposes theirmotion. When reversedVismade large enough, a valueV0 (called stopping or inhibiting potential) is reachedwhen current is reduced to zero. Stopping potentialis that value ofthe retarding potentialdifference betweenthe two electrodeswhich is just sufficient to half themost energetic photoelectrons emitted. As seen fromcurve II of figure, doubling the intensity of the incident light merely doubles the current but does not affect the value ofV0. Now, if vmax is themaximumvelocityofemission of a photoelectron andV0 the stopping potential, then 2max 1 mv 2 = eV0 or vmax = 0 2.eV m

or vmax = 11 2.1.76.10 .V0 = 5.93 × 105 0 V m/s Obviously Emax = eV0 joules = V0 electron-volt If however, the experiment is repeated by varying the frequency of the light, it is found that the stopping potential varies linearlywith frequency as shown in figure. Below threshold frequency, no electrons are emitted, hence stoppingpotential is zero for that reason. But as frequencyis increased above f0, the stopping potentialvaries linearlywith the frequencyofincident light.

ATOMIC PHYSICS www.physicsashok.in 18 Einstein�s photoelectric equationmay be expressed interms of stopping potentialas given below: hf = W0 + 2max 1 mv 2 f0 Frequency Stopping potential, V0 Now, W0 = hf0 and 2max 1 mv 2 = eV0 . hf = hf0 + eV0 or V0 = 0 h(f f ) e. Now, f = c. and f0 = 0 c . V0 = ch e 0 . 1 1 . . . . . . . . = 12.4 × 10�7 0 . 1 1 . . . . . . . . volt . and .0 inmetres . V0 = 12,400 0 . 1 1 . . . . . . . . volt . and .0 in Å C9: Photoelectrons with a maximumspeed of 7 × 105m/sec are emitted froma metal surfacewhen light of frequency 8 × 1014Hz falls on it. Calyculate the threshold frequency of the surface. Sol: Emax = h(f � f0) . 2max 1 mv 2 = h(f � f0) or 12 × 9.1 × 10�31 × (7 × 105)2 = 6.625 × 10�34 (8 × 1014 � f0) . f0 = 4.635 × 1014 Hz C10:Atungsten cathodewhose thresholdwavelength 2300Åis irradiated by ultraviolet light ofwavelength 1800Å. Calculate (i) the maximumenergy of the photoelectrons emitted and (ii) thework function for tungsten, bothin electron-volts. Sol: (i) W0 = 0

12, 400 . = 12, 400 2300 = 5.4 eV (ii) Emax = 12,400 0 . 1 1 . . . . . . . . eV = 12,400 1 1 1800 2300 . . . . . . .= 1.5 eV C11: If light of . = 6000Åfalls on ametalsurface and emits photoelectronswith a velocityof 4 × 105m/s,what is photoelectric thresholdwavelength ?

ATOMIC PHYSICS www.physicsashok.in 19 Sol: K.E. of photoelectrons = 12 × 9.1 × 10�31 × 5 2 19 (4 10 ) 1.6 10. .. = 0.445 eV Energy content of photon of . = 6000Å= 12, 000 6000 = 2.07 eV . W0 = 2.07 � 0.445 = 1.625 eV ..0 = 12, 000 1.625 = 7631 Å C12: Calculate the threshold frequency for gold having photoelectricwork functionequal to 4.8 eV. If light of wavelength 2220Åfalls on gold,what will bemaximumkinetic energyof the photoelectrons coming out? Sol: Energy of the light photon = 12, 000 2220 = 5.58 eV. Out of this, 4.8 eV would be used for dislodging the electron and the balancewould represent its kinetic energy. Emax = 5.58 � 4.8 = 0.78 eV Alternatively, .0 = 12, 400 4.8 = 2583 Å. Hence, wemay use . Emax = 12,400 0 . 1 1 . . . . . . . . eV. C13:When violet light of . = 4000 Au strikes the cathode of a photocell a retarding potential of 0.4 V is required to stop emission of electrons.Calculate (i) light frequency(ii) photon energy(iii)work function (iv) threshold frequencyand (v) net energyafter the electron leaves the surface. Sol: (i) f = c. = 8 10 3 10 4000 10. .. = 7.5 × 1014 Hz (ii) E = hf = 6.625 × 10�34 × 7.5 × 1014 = 4.95 × 10�19 J = 3.1 eV (iii) W0= hf � K.E. = hf � V0 = 3.1 � 0.4 = 2.7 eV (iv) f0 = 0 Wh = 19 34 2.7 1.6 10 6.625 10

.

.

. .

. = 6.5 × 1014 Hz (v) Net energy hf � W0 = 3.1 � 2.7 = 0.4 eV = 6.4 × 10�20 J Example 9:Aphoton ofwavelength 3310Åfalling on a photo cathode ejects an electron of energy 3 × 10�19 J and one of wavelength 5000 Å ejects an electron of energy 0.972 × 10�19 J. Calculate the value of Planck�s constant and the thresholdwavelength for the photo cathode. Sol: hf = W0 + K.E. or hc . = W0 + K.E.

ATOMIC PHYSICS www.physicsashok.in 20 In the first case, 810 h 3 10 3310 10. . . . = W0 + 3 × 10�19 In the second case, 810 h 3 10 5000 10. . . . = W0 + 0.972 × 10�19 Subtracting one fromthe other, h = 6.62 × 10�34 J-s Substituting this value of h,W0 = 3 × 10�19J . .0 = 0 ch W = 8 34 19 3 10 6.62 10 3 10 . . . . . . = 6620 × 10�10 m. vmax = 7.12 × 105 m/s Example 10:Acertainmetalhas a thresholdwavelength of 6525Å. Find the stopping potentialwhen themetal is irradiatedwith (a) monochromatic light having awavelengthof 4000Å. (b) light having twice the frequencyand three times the intensity of that in (a) above. (c) If a material having double the work function were used, what would be the answer to (a) and (b) above? Sol: (a) V0 = 12,400 1 1 4000 6525 . . . . . . . = 1.2 volt (b) Stopping potential is independent ofthe intensity of the incident light but varies directlyas frequencyf provided it ismore than f0. Since frequencyis twice, thewavelength of the light is half i.e. 2000Å V0 = 12,400 1 1 2000 6525 . . . . . . . = 4.3 V (c) Ifwork function is double, then .0 is reduced to half i.e. .0 = 6525 2 =3262.5Å. Since the incident light

has . = 4000 Å, it would not be able to produce photoemission. In the second case, . = 2000 Å. . V0 = 12,400 1 1 2000 3262.5 . . . . . . . = 2.4 volt C14:Acertainmetallic surface is illuminated bymonochromatic light of variablewavelength.No photoelectrons are emitted above awavelength of 5000Å.With an unknownwavelength, a stopping potential of 3.1Vis necessary to stop photoelectric current. Find the unknownwavelength. Sol: Here, .0 = 5000 Å, V0 = 3.1 V, . = ? Now, V0 = 12,400 0 . 1 1 . . . . . . . . . 3.1 = 12,400 1 1 5000 . . . . . . . . . . = 2,222 Å

ATOMIC PHYSICS www.physicsashok.in 21 C15: Light ofwavelength 2000Åfalls on analuminiumsurface. In aluminium, 4.2 eVare required to remove an electron. Determine (i) KE of the fastest emitted photo-electron (ii) KE of the slowest emitted photoelectron (iii) stopping potentialand (iv) cut-offwavelength for aluminium. Sol: Photon energyofthe incident light is = 12400 2000 = 6.2 eV (i) Emax = (6.2 � 4.2) eV = 2 eV = 2 × 1.6 × 10�19 = 3.2 × 10�19 J (ii) Emin = 0 (iii) Ve = Emax = 2 V (iv) .0 = 0 12400 W = 12400 4.2 = 2952.4 Å C16: The stopping potential is 4.6Vfor light offrequency 2 × 1015Hz.When light of frequency 4 × 1015Hz is used, the stopping potential is 12.9V. Calculate the value ofPlanck�s constant. Sol: eV0 = h(f � f0) Substituting the two given values,we get 4.6e = h (2 × 1015 � f0) 12.9e = h(4 × 1015 � f0) Subtracting one fromthe other,we have 8.3e = 2h × 1015 8.3 × 1.6 × 10�19 = 2h × 1015 . h = 19 15 8.3 1.6 10 2 10 . . . . = 6.44 × 10�34 Js Example 11. 10�3Wof 5000Ålight is directed on a photoelectric cell. If the current in the cell is 0.16 µA, the percentage of incident photonswhich produce photoelectrons, is (A) 0.4% (B) .04% (C) 20% (D) 10% Sol. The percentage of incident photonswhichproduce photoelectrons is = n / t 100 N/ t . . . ...(1) . I net . . 16 19 n I 0.16 10 t e 1.6 10 .

.

.

. .

. .

. n 1012 t . . ...(2) and W N hc t . . . . N W 1016 t hc 4 . . . . ...(3) Persentage = 12 16 10 4 100 10 . . [fromeq. (1), (2) and (3)] Persentage . 0.04% Hence (B) is correct.

ATOMIC PHYSICS www.physicsashok.in 22 Example 12. In a photo-emissive cell, with excitingwavelength ., themaximumkinetic energyofelectron isK. If the excitingwavelengthis changed to 34. the the kinetic energyof the fastest emitted electronwillbe : (A) 3K/4 (B) 4K/3 (C) less than 4K/3 (D) greater than 4K/3 Sol. hc . . .K . ...(1) hc K´ 3 / 4 . . . . ...(2) Substracte eqn. (1) fromeq. (2) 4hc hc K´ K 3 . . . . . 4hc 3hc K´ K 3. . . . hc K´ K 3 . . . K K´ K 3 . . K´ K K 3 . . 4K K´ 3 . Hence (D) is correct. Example 13. Let K1 be themaximumkinetic energy of photoelectrons emitted by a light ofwavelength .1and K2 corresponding to .2. If ..1= 2.2, then : (A) 2K1 = K2 (B) K1 = 2K2 (C) K1 < 2 K2 (D) K1 > 2K2 Sol. K.E. = P2 2m P . 2mK P . K h . K . 1K . . 1 2 2 1 K

K . . .

ATOMIC PHYSICS www.physicsashok.in 23 . .1 = 2.2 . 2 2 2 1 2 KK . . . 21 K 4 K . 2 1 K K4 . . 2 1 K K2 . Hence (C) is correct. Example 14. Radiation of two photon energies twice and five times thework function ofmetal are incident sucessively on themetal surface. The ratio of themaximumvelocity of photoelectrons emitted is the two caseswill be (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1 Sol. E1 = 2 . E2 = 5 . E1 = . + K1 2 . = . + K1 K1 = . and E2 = . + K2 5 . = . + K2 K2 = 4 . . 12 K 1 K 4 .2max 1 2max 2 1 mv 1 21 mv 4 2 . 2max1 2max 2 v 1 v 4

. max 1 max 2 v 1 v 2 . Hence (A) is correct. Example 15.When photons of energy 4.25 eVstrike the surface ofametalA, the ejected photoelectrons have maximumkinetic energyTa eVand de-Brogliewavelength.a. themaximumkinetic energyofphotoelectrons liberated from another metal B by photones of energy 4.7 eV is Tb = (Ta � 1.5) eV. If the De-Broglie wavelength of these photoelectrons is .b= 2.a, then find (a) Thework function of a (b) Thework function of b is (c) Ta and Tb

ATOMIC PHYSICS www.physicsashok.in 24 Sol. 2 A Amax h 1 mv 2 . . . . and a A h P . . . A a P . h . . 2 2 A A max 1 P mv 2 2m . wheremismass of electron. . 2A A P h 2m . . . . or 4.25 = 2 A 2a h 2m . . . ...(1) For B, 4.7 = .B + Tb 2 B 2b h 4.7 2m . . . . ...(2) But .b = 2.a and Tb = Ta � 1.5 After solving (a) 2.25 eV (b) 4.2 eV (c) 2 eV and 0.5 eV C17: An isolatedmetal body is illuminatedwithmonochromatic light and is observed to become charged to a steadypositive potential1.0Vwithrespect to the surrounding.Thework function ofthemetal is 3.0 eV. the frequency of the incident light is __________. Sol. h . = . + eV or h . = 3 + 1 = 4 eV .

19 34 4 eV 4 1.6 10 h 6.63 10 . . . . . . . . . = 0.96 × 1015 Hz Example 16. 663mWof light froma 540 nmsource is incident on the surface of ametal. If only 1 of each 5 × 109 incident photons is absorbed and causes an electronto be ejected fromthe surface, the totalphotocurrent in the circuit is _______. Sol. N.t = no. of photon incident per second. . 663 10 3 N hc t . . . . .

ATOMIC PHYSICS www.physicsashok.in 25 . N 663 10 3 663 10 3 t hc 1242 nmeV 540 . . . . . . . . . 9 n / t 1 N/ t 5 10 . . . . . 3 9 9 n 1 N 1 663 10 540 t 5 10 t 5 10 1242 nmeV . . . . . . . . . . . . I ne 5.76 1011 A t . . . . Example 17. Light ofwavelength 330 nmfallingon a piece ofmetal ejects electronswithsufficient energywhich requires voltageV0 to prevent a electron fromreading collector. In the same setup, light ofwavelength 220 nm, ejects electronswhichrequire twich the voltageV0 stop theminreaching a collector. Find the numerical value of voltageV0. (take plank�s constant, h = 6.6 × 10�34 Js and 1 eV= 1.6 × 10�19 J) Sol. 0 1 hc . . . eV . and 0 2 hc . . . 2eV . Here .1 = 330 nm and .2 = 220 nm After solving,0 V 15 volt 8 . Example 18. Asmall 10Wsource of ultraviolet light ofwavelength 99 nmis held at a distance 0.1 mfroma metal surface. The radius of an atomof themetal is approximately0.05 nm. Find (i) the average number of photons striking an atomper second. (ii) thenumber ofphotoelectrons emitted per unit area per secondifthe efficiencyofliberationofphotoelectrons is 1%. Sol. (i) I = intensity = 2 10

4.(0.1) . .w= power incident on atom = . .2 9 2 2 I r 10 0.05 10 4 10 . . . . . . . . . w n hc t . . . . . n w 5 t hc / 16 . . . . .

ATOMIC PHYSICS www.physicsashok.in 26 (ii) no. of photons incident per unit area per second I hc / . . . no. of ejected electrons per unit area per second I 1 1020 hc / 100 80 . . . . . Example 19. The surface ofcesiumis illuminatedwithmonochromatic light of variouswavelengths and the stopping potentials for the wavelengths are measured. The results of this experiment is plotted as shownin the figure. estimate the value ofwork functionof the cesiumand Planck�s constant. 120 �1 �2 0.49 0.5 1.0 1.5 v × 1015Hz supporting potential (volt) Sol. h . = . + eVs or s V he e . . . . Fromgraph 15 h 2 e 0.49 10 . . . 19 15 h 2 1.6 10 0.49 10 . . . . . h = 6.53 × 10�34 Js But 2 e. . . . . . = 2 eV Example 20. In a photoelectric effect set-up a point source of light of power 3.2 × 10�3Wemitsmonoenergetic photons of energy 5.0 eV. The source is located at a distance of 0.8 mfrom the centre of a stationary metallic sphere of work function 3.0 eV and of radius 8.0 × 10�3 m. The efficiency of photoelectron emission is one for every 106 incident photons.Assume that the sphere is isolated and initially neutral and that photoelectrons are instantlyswept awayafter emission. (a) Calculate the number of photoelectrons emitted per second. (b) Find the ratio of thewavelengthof incident light to the de-Brogliewavelength

of the fastest photoelectrons emitted. (c) It is observed that the photoelectronemission stops at a certain time t after the light source is switched on why? (d) Evaluate the time t. Sol. (a) Energyof emitted photons E1 = 5.0 eV = 5.0 × 1.6 × 10�19 J E1 = 8.0 × 10�19 J Power of the point source is 3.2 × 10�3 watt or 3.2 × 10�3 J/s Therefore, energyemitted per second, E2 = 3.2 × 10�3 J. s 0.8 m r = 8.0×10�3m Hence number of photons emitted per second

ATOMIC PHYSICS www.physicsashok.in 27 2 1 1 E n E . or 3 1 19 3.2 10 n 8.0 10 . . . . . n1 = 4.0 × 1015 photons/sec. Number of photons incident on unit area at a distance of 8.0mfromthe source S will be n2 = 1 2 n 4.(0.8) = 4.0 1015 4 (0.64) . . . 5.0 × 1014 photon/sec � m2. The area ofmetallic sphere overwhich photonswill fallis : A = .r2 = .(8 × 10�3)2 m2 . 2.01 × 10�4 m2 Therefore, number of photons incident on the sphere per second are n3 = n2 A= (5.0 × 1014 × 2.01 × 10�4) . 1011 per second But since one photoelectron is emitted for every 106 photons hence number of photoelectrons emitted per second, n = 36 n 10 = 11 6 10 10 = 105 per second or n = 105 per second (b)Maximumkinetic energyof photoelectrons Kmax = Energy of incident photones �work function Kmax = (5.0 � 3.0) eV = 2.0 eV = 2.0 × 1.6 × 10�19 J Kmax = 3.2 × 10�19 J The de-Brogliewavelength of these photoelectronswillbe 1 max h h p 2 K m . . . (Here h = Planck�s constant andm=mass of electron) .

34 1 19 31 6.63 10 2 3.2 10 9.1 10 . . . . . . . . . . .1 = 8.68×10�10 m = 8.68 Å Wavelength of incident light .2 (inÅ) = 112375 E (in eV) or .2 = 12375 5 Å = 2476 Å Therefore, the desired ratio is 21 2475 285.1 8.68 . . . . (c)As soonas electrons are emitted fromthemetal sphere, it gets positively charged and acquires positive potential. The positive potential graduallyincreases asmore andmore photoelectrons are emitted fromits surface. Emissionof photoelectrons is stoppedwhen its potential is equal to the stopping potential required for fastestmoving electrons. (b)As discussed in part (c), emission of photoelectrons is stoppedwhen potential on themetal sphere is equal to the stoppeing potentialof fastestmoving electrons. Since Kmax = 2.0 eV

ATOMIC PHYSICS www.physicsashok.in 28 Therefore, stopping potentialV0 =2 volt.Let q be the charge required for the potentialon the sphere to be equal to stopping potentialor 2 volt. Then . 9 . 3 0 2 1 . q 9.0 10 9 4 r 8.0 10. . . . .. . . q = 1.78 × 10�12 C Photoelectrons emitted per second = 105 [part a] or charge emitted per second = (1.6 × 10�19) × 105 C = (1.6 × 10�14) C Therefore, time required to acquire to charge qwill be 2 14 q 1.728 10 t sec sec 1.6 10. 1.6. . . . or t . 111 second Example 21.Monochromatic radiation ofwavelength .1= 3000Åfalls on a photocelloperating in saturation mode.The corresponding spectral sensitivityof photocellis J = 4.8mA/W.When anothermonochromatic radiation ofwavelength .2 =1650Åand power P = 5mWis incident. It is found thatmaximumvelocityof photoelectrons increases to n = 2 times.Assuming efficiency of photo-electron generation per incident photon to be same for both the cases, calculate (i) thresholdwavelength for the cell and (ii) saturation current in second case. [Given, h = 6.6 × 10�34 Js, c = 3 × 108 ms�1 and e = 1.6 × 10�19 coul.] Sol. Maximumkinetic energy of photoelectrons is given byEk = hc . �W0,where . iswavelength of incident radiation andW0 iswork function of the surface onwhich radiation is incident. . Maximumkinetic energyof photoelectrons emitted byradiation ofwavelength .1 is given by 12 mv12 = 1 hc . �W0 or 21 0 1 mv 2 hc W . . . . . . . . . ...(1) wheremismass of an electron and v1 ismaximumvelocityof photoelectrons. Similarly, for radiation ofwavelength .1, 220

2 1 mv hc W 2 . . . ...(2) But v2 = 2v1, therefore fromequation (2), 21 0 2 2mv . hc .W . ...(3) Fromequations (1) and (3), 0 0 1 2 4 hc W hc W . . . . . . . . . . . or W0 = 3 eV

ATOMIC PHYSICS www.physicsashok.in 29 But work-functionW0 = 0 hc . where .0 is thresholdwavelength. . .0 = 0 hc W = 4125 Å Ans. (i) In saturationmode, spectalsensitivitywithwavelength .1 =3000Åis J = 4.8mA/Wor 4.8mC/J. Itmeans when 1 joule radiation ofwavelength .1= 3000Åis incident, a charge of 4.8mCflows in saturationmode or 4.8 mC e electrons are ejected. Energy of each photon ofwavelength .1 is E1 = 1 hc . . Number of photons in 1 joule radiation ofwavelength .1 1 1 1 E hc . . . No. of electrons ejected by these photons = Je = 3 19 4.8 10 1.6 10 . . . . = 3 × 1016 . Efficiencyof photo-electrongeneration per incident photon, 16 1 3 10 0.0198 ( / hc) . . . . . . Energy of eachphoton ofwavelength .2, E2 = 2 hc . . Rate of incidence of photons ofwavelength .2 in a radiation of power P 2 2 P P E hc .

. . per second Since, efficiency . of photo-electron generation is same for both the case, therefore, rate of ejection of electrons in later case2 . Phc . . . per second . Rate of flow of charge in saturationmode = 2 Phc . . e Cs�1= 13.2 µ Cs�1 But rate of flowof charge is current. Hence, saturation current is second case = 13.2 µA. Ans. (ii) Example 22. A monochromatic point source S radiating wavelength . = 6000Åwith power P = 2watt, an apertureAof radius R = 1 cm and a large screen are placed as shownin fig.Aphotoemissive detector D of surface area S = 0.5 cm2 is placed at centre of the screen. Efficiency of detector for photoelectric emissionper incident photon is . = 0.9. D A L S60cm 6 m (i) calculate photonflux at centre of screen and photo current in the detector. (ii) If a convex lens L of focal length f = 30 cmis inserted in the aperture as shown, calculate new value of photon flux and photo current assuming a uniformaverage transmission of 80%fromthe lens. (iii) Ifwork function of photo-emissive surface isW0 = 1 eV, calculate value ofstopping potential intwo cases (without andwiththe lens in aperture). Given, h = 6.625 × 10�34 J-S, c = 3 × 108 ms�1.

ATOMIC PHYSICS www.physicsashok.in 30 Sol. Photon flux is rate of incidence ofphotons per unit are ofdetector. Therefore, to calculate photon flux, first rate of emission of electrons fromthe source should be calculated. Energy of each photon is E . hc . Rate of emission of energy fromthe source is P = 2 watt = 2 Js�1 . Rate of emission of photons fromthe source is n P P E hc. . . or n = 6.04 × 1018 photons per second (i) Distance of detector fromsource is r1 = 6m . Photon flux at detector, 16 1 2 1 n 1.33 10 4 r . . . . . photons/m2s Rate of incidence of photons on detector = .1 . S Rate of emission of electrons fromdetector = ..1S per second Since, current is charge flowing per second, therefore photo current = (..1S)e = 9.6 × 10�8 amp (ii) When a concave lens is inserted in the aperture, it refracts incident rays. Therefore, photon flux and hence photo-current changes. Distance of lens fromsource is r2 = 0.60m . Photon flux at lens is .´= 2 2 n 4.r = 1.33 × 1018 photons/m2s Considering a very small areaAof the lens, Rate of incidence of photons on this area of lens =A.´ Nowconsidering refractionthrough the lens, u = � 60 cm, f = + 30 cm v = ? Using lens formule, 1 1 1 v u f . . , v = + 60 cm Since, average transmissionfromlens is 80%, therefore, rate of transmissionofphotons fromareaAof lens = 0.8 A.´ But these photons are transmitted in a solid angle subtended by the areaAat P as shown in fig. This solid angle, 2 2 A A v (0.6) . . . . Rate ofphotons transmitted per unit solid angle is T (0.8 A ´) . . . = (0.8 .´) (0.6)2 = 0.288 .´

D S P v 4.80m

ATOMIC PHYSICS www.physicsashok.in 31 Solid angle subtended by unit area of detector at P, .´= 2 1 (4.80) . Photon flux at D, .2 = T.´ = 1.67 × 1016 photons/m2s Rate of incidence of photons on detector = .2S Rate of emission of electrons fromdetector = ..2S . Photo current = (..2S)e = 1.20 × 10�7 amp (iii) Since, stopping potentialV0 is given bye.V0 = 0 . hc .W . .. . .. and . andW0 bothremainunchanged, therefore, stopping potential is same for boththe cases. . 0 0 V 1 hc W 1.07 volt e . . . . . .. . .. PHOTOELECTRIC EFFECT AND WAVE THEORY OF LIGHT According towave theory, when light falls on ametal surface, energy is continuouslydistributed over the surface, energyis continuouslydistributed over the surface.All the free electrons at the surface receive light energy.An electronmaybe ejected onlywhen it acquired energymore than thework function. Ifwe use a low-intensity source, itmaytake hours before an electron acquires thismuch energy fromthe light. In this period, there will be many collisions and any extra energy accumulated so far will be shared with the remainingmetal. Thiswill result in no photoelectron. This is contrary to experimental observations. No matter howsmall is the intensity, photoelectrons are ejected and that too without any appreciable time delay. In the photon theory, lowintensitymeans less number ofphotons and hence less number of electrons get a change to absorb energy.But anyfortunate electrononwhich a photonfalls, gets the fullenergyof the photon andmaycome out immediately. In figure,we illustrate an analogyto thewave the particle behaviour of light. In part (a), water is sprayed froma distance on an area containing several plants. Each plant receiveswater at nearly the same rate. It takes time for a particular plant to receive a certain amount ofwater. In part (b) of the figure,water is filled in identical, loosely-tiedwater bags and a particle physicist throws the bags randomly at the plants.When a bag collideswith a plant, it sprays all itswater on that plant ina very short time. In the sameway,whole of the energyassociatedwith a photon is absorbed by a free electronwhen the photon hits it. (a) (b) Themaximumkinetic energyof a photoelectrondoes not depend on the intensityof the incident light. This fact is also not understood bythewave theory.According to this theory,more intensitymeansmore energy and themaximumkinetic energymust increasewiththe increase inintensitywhichis not true.The dependence ofmaximumkinetic energyonwavelength is also against thewave theory.There should not be anythreshold wavelength according to thewave theory. According to this theory, byusing sufficientlyintense light of any

wavelength, an electronmaybe given the required amount of energyto come out. Experiments, however, showthe existence of thresholdwavelength.

ATOMIC PHYSICS www.physicsashok.in 32 DUAL NATURE OF LIGHT (a) Wave nature: Wave nature of light can be explained on the basis of reflection, refraction, interference, diffraction and polarization. (b) Particle Nature: Energy is transported byenergy particles, photons. It could be explained byphotoelectric effect, Zeeman effect, Comptoneffect etc. MATTER WAVE THEORY OR DE-BROGLIE�S THEORY (a) This theorywas given on the basis of duelnature of light. (b) According to de-Broglie theoryeach and everymoving particle has somewave nature associatedwithitself which is calledmatter waves. (c) Thus,moving particles like e�, proton, neutron, .-particle etc, also behave likewaves. (d) Thesewaves arewaves of probability. (e) The wavelength associatedwith a moving particles is given by . = h/p, where p is themomentumof the particle. (f) Thiswavelength is known as the de-Brogliewavelength of the particle. DE BROGLIE WAVELENGTHS For Photon For moving particle Rest mass Zero m Effectivemass m = 2 Ec = 2 hc. m= 02 2 m 1. v / c Energy(kinetic) E = hv = hc . E = 12 mv2, E = p2 2m Momentum p = Ec = hc. = h. p = mv = 2mE Wavelength . = hp = hc E . = h

p = h 2mE Wavelength for charged particle . = h 2mE accelerated byVvolts K.E. = qV Speed c = 3 × 108 m/s v = 2E /m = 2qV/m De-Broglie�s explanation for stable Bohr�s orbits : (a) De-Broglie suggested that non-radiation ofenergybythe electrons circling in aBohr�s orbit can be explained on the basis on the basis ofthe formation of stationarywaves bythe electrons in circularmotion inBohr�s orbits. . r . O Fourth Bohr-orbit (b) Comparing this to the vibrations ofawire loop suchstationarywaveswould be formed ifeachwave joins smoothlywith the next.

ATOMIC PHYSICS www.physicsashok.in 33 (c) In otherwords the number ofwavelengthsmust be an integer. (d) Condition for stable orbits : An election cancircle around an atomic nucleuswithout radiation energyif the circumference ofits orbit is an integralmultiple ofthe electronswavelength. i.e. 2.r = n... condition for stable orbits. 2.r = n hp . pr = nh 2. . mvr = nh 2. [. p = mv] mvr = nh 2. Asmvr is the angularmomentumof the circling electron. Bohr�s postulate is justified. C18: Calculate the de-Broglie wavelength associated with themotion of earth (mass = 6 × 1024kg) orbiting around the sun at a speed of 3 × 106ms�1. Sol: . = h mv = 34 24 6 1 6.63 10 (Js) (6 10 ) (3 10 )(kg ms ) . . . . . . . . . = 3.68 × 10�65 m NOTE: The wavelengths associated with the motion of macroscopic objects like earth, train etc, are negligible compared to their sizes. This is why the wave-like character of these objects is not observable in our daily life. C19: Calculate the de-Broglie wavelength of an .-particle ofmass 6.576 × 10�27 kg and charge 3.2 × 10�19 coulomb, accelerated though 2000V. Sol: E = kinetic � energy of a-particle = qV . E = 3.2 × 10�19 × 2000 J = 6.4 × 10�16 J deBrogliewavelength, . = h 2mE , [. 2mE =momentumof photon] = 34 27 16 6.63 10 (J s) 2 6.576 10 6.4 10 (J kg) . . . . .

. . . . . = 2.28 × 10�13 m NOTE: The wavelength associated with .-particles is of the order of size of the .-particle. That is why the wave like character of a-particle is observable. C20:Aparticle ofmassmand charge q is accelerated through a potentialdifferenceV. Find (a) its kinetic energy (b)momentum, and (c) de-Brogliewavelength associatedwith itsmotion. Sol:When the particle is accelerated througha potentialdifferenceV, gain in kinetic energyis given byK= qV. (a) Thus, kinetic energy,K= qV. (b) Momentumof the particle (p) is given byK= p2 2m p = 2mK = 2mqv . Momentum= 2mqv

ATOMIC PHYSICS www.physicsashok.in 34 (c) De- Brogliewavelengthis given by, . = hp = h 2mqV Thus, wavelength= h 2mqV Example 23. Assume that the de-Broglie was associatedwith an electron can forma standing wave between the atoms arranged in a one dimensional arraywith nodes at each of the atomic sites. It is found that one such standingwave is formed if the distance d between the atoms of the array is 2 Å.Asimilar standing wave is again formed if d is increased to 2.5 Åbut not for any intermediate value of d. Find the energy of the electron in eV and the least value of d for which the standing wave of the type described above can form. Sol. Fromthe figure it is clear that p . (./2) = 2 Å (p + 1) . ./2 = 2.5 Å . ./2 = (2.5 � 2.0) Å = 0.5 Å or ..= 1 Å = 10�10 m (i) deBrogliewavelengthis given by 2 Å2.5 Å N N p-loops (p + 1) loops /2 h h p 2 km . . . K= kinetic energy of electron 2 34 2 17 2 31 10 2 K h (6.63 10 ) 2.415 10 J 2m 2(9.1 10 )(10 ) . . . . . . . . . . . 17 19 K 2.415 10 eV 1.6 10 . . . . . . . . . . . . K = 150.8 eV (ii)N N The least value of d will bewhen only one loop is formed

. dmin = ./2 or dmin = 0.5 Å C21: Find the de-brogliewavelength associatedwith an electron accelerated through a potential difference of 30 kV. Sol: Kinetic energyof the electron K = qV = e(30 kV) = 30 keV = 30 × 103 × 1.6 × 10�19 J = 4.8 × 10�15 J Now,momentumof the electron = 2mK = 2.9.1.10.34 . 4.8.10.15 kg.J = 2.96 × 10�24 kg.m/s . De-brogliewavelength, . = h Momentum = 34 24 6.63 10 J s 2.96 10 Kgm/ s . . . . . = 2.24 × 10�10 m = 2.24 A

ATOMIC PHYSICS www.physicsashok.in 35 ATOMIC STRUCTURE ATOMIC MODEL By now, it iswell-known thatmatter, electricity and radiation etc. are all atomic incharacter.Although, no one has so far seen individualatoms, there is no doubt that they reallyexist. In 1895, it was discovered by J.Perrin in Paris that the cathode rays consist of negatively-charged practices called electrons. In 1897, J.J. Thomson measured the e/mratio for an electronwhereas its chargewasmeasured byMillikan in 1906 byhis famous oil-drop experiment.Mass of the electronwas found by dividing charge e bythe ratio e/m.Discoveryofpositive rays during the latter part of 19th centuryindicated that a normalatomconsisted of both negative and positive charges. But howthese charges are distributed in an atomwas not known at that time. Consinuous efforts have beenmade since then to studythe physical structure of anatomsuch as its extra-nuclear electronic structure chieflywiththe help of spectralproperties of atoms.To account for the spectroscopic dataobtained experimentallyover the years, several theories regarding atomic structure have been proposed fromtime to timewhich are called the atomicmodels.Various atomicmodels proposed by scientists over the last fewdecades are: (i)Thomson�sPlumpuddingmodel, (ii) Rutherford�sNuclearmodel, (iii) Bohr�smodel(iv) Sommerfeld�s Relativisticmodel (v)Vectormodel and finally(vi)Wave-mechanicalmodel. These differentmodels have been suggested one after the other inaneffort to get a satisfactoryinterpretation of the experimentaldatawhich, it is hoped, willultimately lead to a perfect and complete understanding of the physical structure of an atom. Thomson�s Plum PuddingModel According to thismodel, the atomis regarded as a heavy sphere of positive charge seasonedwith enough electron �plums� to make it electricallyneutral. Thomson visualized the positive charge ofanatombeing spread out uniformly throughout a sphere of about 10�10 metre radius with electrons as smaller particles distributed in circular shells as shown infigure.Whereas the net force + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + ++ + + + + + +

+ + + + + + + + + + + + + + + + + + + � � � � � � exerted bythe positively-charged sphere on each electron is towards the centre of the sphere, the different electrons experiencemutual repulsion and get arranged in the formof circular shells. This atomicmodelwas given up after some time because it could not provide any satisfactorymechanism for explaining the large deflection suffered by.-particles inRutherford�s experiment. Rutherford�s Experiment on .-particle Scattering As showninfigure high-speed a-particle (i.e. heliumnucleieachwitha charge of+2e) fromsome radioactive material like radiumor radon, confined to a narrowbeamby a hole in a lead block, weremade to strike a verythin goldfoilG.Whilemost ofthe .-particleswent straight throughthe foilas iftherewerenothing there (and produced scintillations on a fluorescent screen), some of them�collided�with the atoms ofthe foil and were scattered around at various angles -a fewbeing turned back towards the source itself. 150º 45º 60º . .. . . . Radium Lead block + Beam of .-particles

ATOMIC PHYSICS www.physicsashok.in 36 Althoughsome smallangle scattering could be expected fromThomson�smodel, large-angle scatteringwas absolutelynot expectedat all. Further detailedexperiments inRutherford�s laboratorybyGeiger andMarsden showed conslusivelythat large-angle scattering canbe expected onlyifone assumes that amassive positive point charge exists at the centre of each gold atomas shown in figure. According to thismodelproposed byRutherford in 1911, the positivemassive part of anatomis assumed to be concentrated in a very smallvolume at its centre. This central core, nowcalled nucleus, is surrounded by a cloud of electronswhichmakes the entire atomelectrically neutral. . . . P+ZeNucleus Asymptote of trajectory Trajectory of .-particle The large-angle scattering of positively-charged .-particles could be easily explained on this atomic model as shown in figure. This scattering is due to themutual repulsion (as per Coulomb�s law) between the .-particles and the concentrated positive charge on the nucleus. The .-particle approaches the positively-charged nucleus alongAO. If there were no repulsion fromthe nucleus, it would have passed at a distance of p fromit.However, due to coulombic force ofrepulsion, the .-particle follows a hyperbolawith nucleus as its focus.The linesAOand A.Oare the asymptotes of the hyperbola and present approximatelythe initial and finaldirections of the .- particlewhen it has passed out of the effective range of the nuclear electric field. As seen, the .-particle is deflected through an angle .. The perpendicular distance fromthe nucleus to the lineAOis called the impact parameter and is denoted by p.The Rutherford�s scattering formula is tan 2. = 1 2 Q Q 2pE where Q1 = Charge of the incoming .-particle Q2 = Charge ofthe scattering nucleus E = Kinetic energy of the incident .-particle p = Impact parameter Distance of Closest Approach Suppose that an .-particle approaches a positively-charged nucleus for a head on collisionwith a kinetic energy ofK.As shown in figure at pointA, the repulsive force ofthe nucleus is so strong as to stop the .- particlemomentarily.At this point, all the kinetic energyof the .-particle is converted into potential energy. Let Dbe the distance of closest approach of the .-particle. The potential at pointAdue to nuclear charge Ze is

= 0 Ze 4.. D + D A . Potential energy of the .-particlewhen at pointAis = 0 Ze.2e 4.. D = 2 0 2Ze 4.. D . K = 2 0 2Ze 4.. D or D = 2 0 2Ze 4.. K Since .-particles are generally obtained fromnatural radioactive substances, their kinetic energyK is known. Hence, value ofDcan be found easily fromthe above equation.

ATOMIC PHYSICS www.physicsashok.in 37 C22: InGeiger-Marsden experiment on .-particle scattering fromgold foil, the kinetic energy of .-particles usedwas 7.68MeV. Calculate the distance of closest approach of .-particle if atomic number of gold in 79. Sol: D = 2 0 2Ze 4.. K Here, Z = 79, e = 1.6 × 10�19 C; .0 = 8.854 × 10�12 F/m K = 7.68 MeV = 7.68 × 1.6 × 10�13 J D = �19 2 12 13 2 79 (1.6 10 ) 4 8.854 10. 7.68 1.6 10. . . . .. . . . . = 2.96 × 10�14 m Major Deficiencies in Rutherford�s Nuclear Model It was found later on that Rutherford�smodel had two serious drawbacks concerning (i) distribution of electrons outside the nucleus and (ii) the stabilityof the atomas awhole. It canbe shown that electrostatic forces between the positive nucleus and the static negative extra-nuclear electrons are not enough to produce equilibriumin such a nuclear atom. For example, consider the case of an atom* having two electrons and a nucleuswith a charge of+2e. Suppose the electrons are symmetrically placed at a distance of r fromthe nucleus and are stationary.The force of attraction between the nucleus and each of the electrons is F = e. 2 2e 4. .r = 2 2 2e 4..r while the force of repulsion between the two electrons is 2 2 e 4...4r = 2 2 e 16..r . Since the force of attractionis eight times the force of repulsion, the electronswill fall into the nucleus therebydestroying the stable structure of the atom. �� + + + + 2e + + + + � 2e �

v v r me + � Photon (a) (b) (c) 2e (hv) To overcome this difficulty, Rutherford suggested that stability can be achieved (as in a solar system) by assuming that electrons, instead ofbeing static, revolve round the nucleuswithsucha speedthat the centrifugal force balances the attractive force exerted bythe nucleus onthe electrons.As seen fromfigure (b), condition for stabilityis achievedwhen mv2 r = 2 0 2e.e 4.. r or mv2r = 2 0 2e 4. . In general, ifZ is the atomic number, then nuclear charge isZe, so that the above relation becomes mv2 r = 2 0 Ze.e 4.. r or mv2r = 2 0 Ze 4. .

ATOMIC PHYSICS www.physicsashok.in 38 Incidentally, itmaybe noted that according to the above relation, it is possible to have an infinite number of orbits inwhich electrons can rotate. But this assumptionofrevolving electrons ledto serious difficultyfromthe point ofviewo the electromagnetic theoryaccording towhich an accelerated chargemust continuouslyemit electromagnetic radiation or energy. Since an electronrevolving in a circular orbit has centripetal acceleration (= v2/r), itmust radiate energy as per the laws of classical electrodynamics. Due to this continuous loss of energy, the electronswillgraduallyapproachthe nucleus bya spiral path and finally fall into it as shown infigure (c). Hence, it is seen that the orbitalmotion of the electron destroys the verypurpose forwhich itwas postulatedi.e. the stabilityofthe atom.Obviously, eitherRutherford�s nuclear atomicmodelwithrevolving electrons isdefective or the classicalelectromagnetic theoryfails inthis particular case. This dilemmawas solved in 1913 byNeils Bohrwho proposed an improved version ofRutherford�s atomicmodel. Bohr�s Atomic model Thismodel (first proposed for hydrogen atombut later applied to other atoms as well) retains the two essential features ofRutherford�s planetarymodel i.e. (i) the atomhas amassive positively-charged nucleus and (ii) the electrons revolve round their nucleus in circular orbits the centrifugal force being balanced, as before, by the electrostatic pull between the nucleus and electrons. However, he extended thismodel further byutilizing Planck�sQuantumTheory. Hemade the following three assumptions: (iii) an electron cannot revolve round the nucleus in any arbitrary orbit but in just certain definite and discrete orbits.Onlythose orbits are possible (or permitted) forwhichthe orbital angularmomentum(i.e.moment of momentum) ofthe electron is equal to anintegralmultiple of h 2. i.e. orbitalangularmomentum= nh 2. where n is an integer and h is Planck�s constant. Such orbits are also known as stationary orbits. (iv) while revolving in these permitted stationary (or stable) orbits, the electron does not radiate out any electromagnetic energy. Inother words, the permissible orbits are non-radiating paths of the electron. (v) theatom radiatesout energy only when an electron jumps from oneorbit to another. If E2 and E1 are the energies corresponding to two orbits before and after the jump, the frequencyof the emitted photon is given bythe relation E2 � E1 = hv or .E = hv where v is the frequency of the emitted radiations. C 23: If I is the moment of inertia of an electron and . its angular velocity, then as per assumption (iii) given above

.I = nh 2. or (mr2).= nh 2. or (mr2 )v r = n.h 2. or mvr = n.h 2. Alternatively, sincethemomentumoftherevolvingelectronismv, itsmoment about the nucleus is =mvr Hence mvr = n.h 2. .....(i) e� m v +ze r

ATOMIC PHYSICS www.physicsashok.in 39 where, n= 1, 2, 3 for the first second and third orbits respectively. It is called the principalquantumnumber and because it can takewhole number values only, it fixes the sizes ofthe allowed orbits (also calledBohr�s circular orbits). r1 r2 r3 n = 1 n = 2 n = 3 +ZeE1 E2 E3 Permitted orbits (a) KLM +ZeE1 Electron Jump (b) E2 .E E3 Let the different permitted orbit have energies ofE1, E2, E3 etc. as shown in figure (a). The electron can be raised from n = 1 orbit to any other higher orbit if it is given proper amount of energy.When it drops back to n =1 orbit after a short intervalof time, it gives out the energydifference .E in the formof a radiation as shown in figure (b). The relationbetween the energy released and frequency of the emitted radiation is E2 � E1 = hv or .E = hv Expressions for velocity, radius, energy of electron and orbital frequency in Bohr�s orbit Here it should be kept inmind that Bohr�smodelis valid onlyfor hydrogen atomand hydrogen-like ions. In otherwords,we can saythat is applicable to hydrogen atomand ions having just one electron. Examples of such ions are He+, Li++, Be+++ etc. According to Bohr�s + ze r �e, m First postulate, v 2 2 0 Ze 4.. r = mv2 r where 0 1 4. . = 9 × 109 Nm2 C�2 v = velocityof electron and .0 = 8.854 × 10�12 C2/Nm2 m = mass of an electron = Absolute permittivity Z = atomic number

of vacuumor free space r = radius ofthe orbit e = magnitude of charge on an electron n = principalquantumnumber i.e. orbit number . r = 2 2 0 Ze 4.. mv .....(i) Fromfourth postulate ofBohr, we have mvr = nh 2. .....(ii) where, n = 1, 2, 3, 4, ........ i.e., n is a positive integer. Fromequations (i) and (ii),we get mv × 2 2 0 Ze 4.. mv = nh 2. . v = 2 0 ze 2.. nh Now, let us substitute the value of v in equation (ii),

ATOMIC PHYSICS www.physicsashok.in 40 m × 20 ze r 2 nh . . = nh 2. 2 2 0 2 r n h mZe . . . Note that for fixed n, v . Z, r . 1Z For fixed z, v . 1n , r . n2 IfV1 is the speed of the electron in the 1st orbit Then, Vn = V1 Zn , where V1 = c 137 , where c is speed of light. If a0 = first Bohr radius = 0.53Å, then rn = a0 n2 Z K.E. of the electron = 12 mv2 = 2 4 2 2 2 0 mZ e 8. h n = K = kZe2 2r , Potential energy of the atom= � 20 Ze 4.. r = U = � kZe2 r [k = 1/4...0] U = � 2 4 2 2 2 0 mZ e

4. h n . Total energy of the atom, E = K + U E = � 2 4 2 2 2 0 mZ e 8. h n In general, En = �13.6 22 Zn in eV. Orbital frequency for the electron, v = V 2.r = 2 2 2 2 0 0 Ze mZe 2 2 hn h n . . .. . .. . v = 2 4 2 3 3 0 mZ e 4. n h Time period of revolution (T) is given by T = T1 32 nZ where T1 = time period of revolution in the 1st orbit = 1.52 × 10�16 s = 0 1 2 a v. NOTE : It is assumed that the acceleration of the nucleus is negligible on account of its large mass. Some important results for H-atoms when n = 1 1. Bohr radius, a0 = 0.53 Å rn = a0n2/Z 2. v1 = 2.18 × 106 ms�1 . c/137 3. E1 = �13.6 eV = 1 U2

ATOMIC PHYSICS www.physicsashok.in 41 4. K1= 13.6 eV = � 1 U2 5. U1= �27.2 eV 6. v1 = 6.6 × 1015 Hz 7. 1 rydberg = �13.6 eV 8. B1 = 12.5 tesla, where B1 is themagnetic field at the centre ofBohr atomdue to the current generated by themotion ofelectron in 1st orbit. Ground state and excited states The state ofan atomwith the lowest energyis called its ground state or normal state. For ground state, n= 1. The stateswith higher energies are called excited states. For the first excited state, n= 2; for the 2nd excited state, n = 3 and so on. Formth excited state, n =m+ 1 Ionization energy and ionization potential Theminimumenergyneeded to ionize an atomis called ionization energy. The potentialdifference throughwhich anelectron should be accelerated to acquire the value of ionization energyis called ionization potential. The value of ionization energyofH-atomin ground state is 13.6 eVthat of ionization potential is 13.6 eV. Binding energy Binding energyof a systemis the energyneeded to separate its constituents to large distances or itmay be defined as the energy releasedwhenits constituents are brought frominfinityto formthe system. The value ofbinding energyofH-atomis 13.6 eV, identical to its ionization energy. Excitation energy and excitation potential The energyneeded to take the atomfromits ground state to an excited state is called the excitation energy of that excited state. The potentialdifference throughwhich an electron should be accelerated to acquire the value of excitation energyis called excitation potential. NOTE : (A) I.E. = E. � E1 = �E1 = Binding energy of theH-atom. (B) Movement of electron in circular orbits in aBohr atomcauses electric current inthe orbit. This current will lead to self-generatedmagnetic field in the atomand alsomagnetic current (m) (a)Magnetic field(B) IfBis themagnetic field generated at the centre of atom, then H-atom +e a0 �e i B = 00i 2a . , where i is the current due to motion of the electron and a0 is the 1st Bohr-radius. Hence, i = ev

hence B = 0 0 e 2a . .

ATOMIC PHYSICS www.physicsashok.in 42 Putting the values of v and a0, we get B = 12.5 Tesla (b) Magneticmovement vector ( .. ) .. = iA . . . = iA = .ev 20a (c) Relation between .. & L . (angularmomentumvector) µ = iA = .ev 20a , [i = ev, A= . 20a ] L = mV1a0 = m2.v 20a [V1 = 2.va0] . L. = 20 20 e a m2 a . ... = e 2m . L. = e 2m Vectorially, .. = e 2m . L . Example 24: The quantumnumber ofBohr orbit inH-atomwhose radius is 0.01mmis (a) 223 (b) 435 (c) 891 (d) none of these Sol: (b)We know that rn = a0n2 n2 = n0 ra n = 310 0.01 10 0.529 10.. .. . n = 435 Example 25:The quantumnumber n in theBohr�smodel ofH-atomspecifies: (a) radius of the electron (b) energyof the electron

(c) angularmomentumof the electron (d) allof these Sol: (d) rn = 2 0 2 h me .. . . . . . . × n2 En = � 4 2 2 2 0me 8. h n , Ln= n h 2 . . . . . . . Example 26: The radius Bohr�s orbit in ground state for H-atomis (Take .0 = 8.86 × 10�12 C2/Nm2, h = 6.6 × 10�34 J-s) (a) 0.528 Å (b) 0.0528 Å (c) 5.28 × 10�10 m (d) 5.28 × 10�10 cm Sol: (a) . a0 = 2 0 2 h me .. (in this case, Z = 1, n = 1) a0 = 34 2 12 31 19 2 (6.6 10 ) (8.86 10 ) 3.14 (9.1 10 ) (1.6 10 ) . . . . . . . . . . . = 0.528 Å

ATOMIC PHYSICS www.physicsashok.in 43 Example 27: The speed of the electron inthe first Bohr orbit ofH-atomis (Take c = speed of light in vacuum) (a) c (b) c 13.6 (c) c 137 (d) 137 c Sol: (c) For H-atom,we knowthat v = 20 e 2. h [since n this case, Z = 1 and n = 1] = 20 e c 2. hc But 2 0 e 2. hc = 1 137 = fine structure constant . v = c 137 Example 28: In a H-atom, binding energy of the electron in the ground state is E1. Then the frequency of revolution of the electron in the nth orbit is (a) 1 3 2E n h (b) 3 1 2E n h (c) 1 3 2mE n h (d) none of these Sol: (b) The frequencyof revolution of the electron in the nth orbit is given by v = 2 4 2 3 3 0 mZ e 4. h n = 13 2E hn where E1 = 2 4 2 2 0

mZ e 8. h C24: Calculate the energy of aHe+ ion in its first excited state. Sol: . En =� 2 2 (13.6eV)Z n Here, Z = 2, n = 2 . En = 2 2 13.6 2 n . . eV = �13.6 eV C25. An electron in a hydrogen like atomis in an excited state. It has a total energy of �3.4 eV. Calculate : (i) the kinetic energy (ii) the de-Brogliewavelength of the electron Sol. (i)Kinetic energyof electron in the orbits of hydrogen and hydrogen like atoms = | Total energy | . Kinetic energy = 3.4 eV (ii) The deBrogliewavelength is given by h h P 2 Km . . . (K= kinetic energyof electron) Substituting thevalues,we have 34 19 31 (6.6 10 J s) 2(3.4 1.6 10 J)(9.1 10 kg) . . . . . . . . . . = 6.63 × 10�10m or . = 6.63 Å

ATOMIC PHYSICS www.physicsashok.in 44 Example 30. The electron in a hydrogen atommakes transition fromMshell to L. The ratio ofmagnitudes of initial to final centripetal acceleration ofthe electron is (A) 9 : 4 (B) 81 : 16 (C) 4 : 9 (D) 16 : 81 Sol. . rn × n2 and vn × 1n2n n n v a r . n 2 2 a 1 n n . . n 4 a 1 n . n 4 a Kn . M 3 a a K81 . . {. M = 3} L 2 a a K16 . . {. L = 2} ML a 16 a 81 . Hence (D) is correct. Example 31. The angularmomentumopf an electron in the hydrogenatomis 3h 2. .Here h is Planck�s constant. The kinetic energyof this electronis : (A) 4.53 eV (B) 1.51 eV (C) 3.4 eV (D) 6.8 eV Sol. L nh 3h 2 2 . . . . . n = 3 In the electronic third orbit, the energy of electron 22 Z E 13.6

n . E 13.6 19 . . E 13.6 1.51 eV 9 . . Hence (B) is correct. Example 32. A particle of mass m moves along a circular orbit in a centrosymmetrical potential field U(r) = kr2 2 .Using theBohr�s quantization condition, find the permissible orbital radii and energy levels of that particle.

ATOMIC PHYSICS www.physicsashok.in 45 Sol. F = � dU dr = � kr Negative signimplies that force is acting towards centre. The necessary centripetal force to the particle is being provided bythis force F. Hence mv2 r = kr ...(1) and mvr = n h where h h 2 . . . .. . .. ...(2) solving equations (1) and (2) we get r = rn = n h m. where . = km and total energy E = U + K= kr2 2 + 12 mv2 Substituting the values,we get E = n h . = En Example 33. In a hypothetical systema particle ofmass mand charge �3q is moving around a very heavy particle having charge q.AssumingBohr�smodel to be true to this system, the orbital velocity ofmassm when it is nearest to heavyparticle is (A) 2 0 3q 2. h (B) 2 0 3q 4. h (C) 0 3q 2. h (D) 0 3q 4. h Sol. 2 e mv F r . 2 2 0

q 3q mv 4 r r . . .. . . 20 3 q mvr v 4 . .. . mvr h 2 . . . . . 20 3 q h v 4 2 . .. .2 0 v 3 q 2 h . . Hence (A) is correct.

ATOMIC PHYSICS www.physicsashok.in 46 SPECTRUM Dispersed light arranging itself ina patternofdifferent wavelength is referred to as a spectrum. Light coming froma sourcemay be dispersed bya prismor by any other dispersingmedium. Whenwhite light falls on a prismand the transmitted light is collected on awhitewallorwhite paper then a spectrumis obtainedwhich consists ofdifferent colours fromred to violet. Kinds of spectra : (A) Emission spectra:When a light beamemitted by certain source is dispersed to get the spectrum, it is called an emission spectrum. An emission spectrumcan be three types : (a) Continuous spectrum: That emission spectrumwhich is obtained by continuouslyvaryingwavelength, is called continuous emissionspectrum. In this case,when light is dispersed, a bright spectrumcontinuously distributed ona dark background is obtained. Light emitted froman electric bulb, a candle or a red hot iron piece comes under this category. (b) Line spectrum: The atoms and molecules can have certain fixed energies.An atomormolecule, in an excited state, can emit light to lower its energy. Light emitted in such a process has certain fixedwavelengths.When such a light is dispersed, certain sharp bright lines on a dark background is obtained. Such a spectrumis called line emission spectrum. Atomic energy levels Energy Line spectrum . For example, when electric discharge is passed through sodiumvapour, they vapour emits light of the wavelength 589.0 nmand 589.6 nm.Whenthis light is dispersed bya high resolution grating , one obtains two bright yellowlines on a dark background. (c) Band spectrum: Thewavelengths emitted bythemolecular energylevelswhich are generallygrouped into severalbunches, are also grouped; eachgroup beingwell separated fromthe other.The spectrumlooks like separate bands of varying colours. Such a spectrumis called band emission spectrum. Molecular energy levels Energy Band spectrum . (B) Absorption spectrum:Whenwhite light is passed throughan absorbingmaterial, thematerialmayabsorb certainwavelengths selectively.When the transmitted light is dispersed, dark lines or bands at the positions of themissing (absorbed) wavelengths are obtained. Such type of spectrumis called absorption spectrum. White lightAbsorbing material Absorption spectrum

ATOMIC PHYSICS www.physicsashok.in 47 An absorption spectrum may be of two types : (a) LineAbsorption spectrum : Light may be absorbed by atoms to take themfromlower energy states to higher energystates. In the similarwaywhenwhite light is passed through a gas, the gas is found to absorb light of certainwavelength. The absorption spectrumconsists of dark lines on bright background. Such a spectrumis calleda line absorption spectrum.When light coming fromthe sun is dispersed, it shows certain sharply defined dark lines. This shows that certainwavelengths are absent. There missing lines are called Fraunhofer lines. (b) Band Absorption spectra : If absorbing media is polyatomic such as H2, CO2 or KMnO4 solution, instead of dark lines we get few characteristic dark bands (against coloured background) called band absorption spectra. Hydrogen Spectra : If hydrogen gas enclosed in a sealed tube is heated to high temperature, it emits radiation. This radiation consists of components of different wavelengths which deviate by different amounts. The radiationwith different amounts ofdeviation formsH-spectrum. Explanation of hydrogen spectra by Bohr (a) The electron in aH-atomif not disturbed remains in the ground state (i.e. n = 1 state).When the electron receives energyfromoutside, it is elevated to anyone of the higher permitted states (excited state). (b) The electronremainsonlyfor a short intervaloftime (generallyin theorder of10�8 s) in the excited state and comes back to the ground state finally. (c) The electroncan reach the ground state fromanyone of the excited states inmanyways.As a result,many electron transitions take place. (d) According toBohr, all electron transitions terminating at a particular state give rise to a particular spectral series. Series limit Energy Lyman series Balmer series Paschen series Brackett series n =1 n = 2 n = 3 n = 4 n = 5 n = . E = 0 (e) Lyman: nf = 1: 1. = � 1 Ech 2 2 1 1 1 n

. . . . .

. . n = 2, 3, 4,.......... Balmer: nf = 2: 1. = � 1 Ech 2 2 1 1 2 n . . . . . . . n = 3, 4, 5, .......... Paschen: nf = 3: 1. = � 1 Ech 2 2 1 1 3 n . . . . . . . n = 4, 5, 6, .......... Brackett: nf = 4: 1. = � 1 Ech 2 2 1 1 4 n . . . . . . . n = 5, 6, 7, ......... Pfund: nf = 5: 1. = � 1 Ech 2 2 1 1 5 n . . . . . . . n = 6, 7 , 8,......

ATOMIC PHYSICS www.physicsashok.in 48 (f) If an electronmakes a jump fromthe nith to nfth orbit (ni> nf), the extra energyEi � Ef is emitted as a photon of electromagnetic radiation. The correspondingwavelength is given by 1. = i f E E hc . where c = speed of light in vacuum. According to Bohr,we canwrite 1. = 2 4 2 3 0 mZ e 8. ch 2 2 f i 1 1 n n . . . . . . . 1. = RZ2 2 2 f i 1 1 n n . . . . . . . where R = 4 2 3 0 me 8. ch is called theRydberg constant. (g) The value of R is 1.0973 × 107m�1 NOTE: � En = 2 2 RhcZ n . , 1 rydberg = �13.6 eV, Rhc = 13.6 eV (h) 1. is calledwave number ( .. ) of the line and 2. . is called angular wave number of the line. (i) Photon energy= Ep = hv (j) Momentumof a photon = p = Photon Energy speed of light . p = p Ec Important points regarding H-spectra

(a) The sharplydefined, discretewavelengths exist in the emitted radiation in theH-spectrum. (b) Ahydrogen sample emits radiationwithwavelengths less than those inthe visible range (i.e. uv light) and alsowithwavelengthsmore than those in the visible range (i.e. infrared). (c) the linesmay be grouped in separate series. (d) In each series, the separation between the consecutive wavelengths decreases as wemove fromhigher wavelength to lowerwavelength. (e) Aparticularminimumwavelength in each series approach a limiting value knownas series limit. (f) The series corresponding to uv region, visible region and infrared region are known asLyman, Balmer and Paschen series respectively. (g) In the Balmer series of hydrogen, theH. line (3 . 2) is red, theH. line (4 . 2) is blue, theH. ( 5 . 2) and H.(6 . 2) lines are violet, and the other lines are in the near ultraviolet (uv). (h) Theoretically possible no. of emission spectral lines = n(n 1) 2. . (i) Theoretically possible no. of absorption spectral lines = (n � 1)

ATOMIC PHYSICS www.physicsashok.in 49 (j) Approximate range ofwavelength for different colours of visible light Colour Wavelength Range Violet + Indigo 3800 Å to 4500 Å Blue 4500 Å to 5000 Å Green 5000 Å to 5500 Å Yellow 5500 Å to 6000 Å Orange 6000 Å to 6500 Å Red 6500 Å to 7200 Å Infrared rays: 720 nmto 50 mm Ultraviolet light: 10 Å to 3800 Å Limitations of Bohr�s Model (a) It is valid onlyfor one electron atomand hydrogen-like ions e.g. : H, He+, Li+2, Na+10 etc. (b) Orbitswere taken as circular but according to SOMMER field these are elliptical. (c) Intensity of spectral lines could not be explained. (d) Nucleuswas taken as stationary but it also rotates onits own axis. (e) It could not be explained theminute structure in spectrumline. (f) This does not explain the ZEEMANeffect (splitting up of spectral lines inmagnetic field)&Stark effect (splitting up in electric field). (g) This does not explain the doublets in the spectrumof some of the atoms like sodium(5890Åto 5896Å). C26. Total number of emission lines fromsome excited state n1 to another energy state n2(< n1) is given by 1 2 1 2 (n n )(n n 1) 2 . . . . For example total number of lines fromn1 = n to n2 = 1 are n(n 1) 2. . C27. As the principalquantumnumber n is increased in hydrogenand hydrogen like atoms, some quantities are decreased and some are increased. The table given belowshowswhich quantities are increased and which are decreased. Table Increased Decreased Radius Speed Potentialenergy Kinetic energy Totalenergy Angular speed Time period Angularmomentum C28. Whenever the force obeys inverse square law 2 F 1r . . . .. .. , and potential energy is inversely proportional to r, kinetic energy(K), potential energy(U) and total energy (E) have the following relationships. K = | U | 2 and E = �K = U2 . If force is not proportional to 2

1r or potential energyis not proportional to 1r , the above relations do not

ATOMIC PHYSICS www.physicsashok.in 50 hold good. In JEE problems, this situation arises at two places, in an atom(betweennucleus and electron) and in solar system(between sun and planet). C29. Total energy of a closed systemis always negative and the modulus of this is the binding energy of the system. For instance, suppose a systemhas a total energyof �100 J. Itmeans that this systemwill separate if 100 J of energyis supplied to this.Hence, binding energy of this systemis 100 J. Thus, totalenergyof an open systemis either zero or greater than zero. C30. Kinetic energyofa particle can�t be negative,while the potentialenergycan be zero, positive or negative. It basically depends on the reference point where we have taken it zero. It is customary to take zero potential energywhenthe electron is at infinite distance fromthe nucleus. In some problemsupposewe take zero potential energyin first orbit (U1 = 0), then themodulus of actual potential energyin first orbit (when reference point was at infinity) is added inUand E inall energy states, whileKremains unchanged. Example 33:The value of series limit inLyman series is: (a) 121.6 nm (b) 91.2 nm (c) 656.3 nm (d) 365.0 nm Sol: min 1 . = R.1. 1 . .. ... min 1 . = R[1 � 0] .min = 7 1 1.097.10 = 91.2 nm C31: Find the longest wavelength present in theBalmer series of hydrogen: Sol: In the Balmer series, nf = 2. The longest wavelength in this series corresponds to the smallest energy difference between energy levels. Hence the initial statemust be ni = 3. . 1. = R 2 2 f i 1 1 n n . . . . . . . = R 2 2 1 1 2 3 . . . . . . . = R 1 1 4 9 . . . . . . . 1. =

5R 36 . = 7 36 5.1.097.10 = 6.56 × 10�7 m . = 656 nm(near the red end of the visible spectrum) C32:Ahydrogen atomemits uv radiation of102.5 nm. Calculate the quantumnumbers of the states involved in the transition. Sol: The uv radiation 102.5 nmlies inthe lyman region of spectrum. Thus nf=1 . 1. = R 2i 1 1 n . . . . . . . . 2i 1 n = 1 � 1.R . 2i 1 n = 1 � 9 7 1 102.5.10. .1.097.10 = 1 � 1 1.124

ATOMIC PHYSICS www.physicsashok.in 51 2i 1 n = 0.1 2i n = 10 ni . 3 Hence transition is from3 . 1. C33:What is the unit of reciprocalofRydberg constant in S.I. units ? Sol:We know that 1. = R 2 2 f i 1 1 n n . . . . . . . the terminthe bracket is unitless. Now, we canwrite 1. = R . 1R = . Hence, the unit of 1R willbemetre i.e.m. C34: Howmanydifferent wavelengthsmay be observed in the spectrumfroma hydrogen sample if the atoms are excited to stateswith principal quantumnumber n ? Sol: The totalnumber of possible transitions is (n � 1) + (n � 2) + (n � 3) + ............. + 2 + 1 = n(n 1) 2. C35: Consider the following two statements: (A) Line spectra contain information about atoms only (B) Band spectra contain information aboutmolecules (a) BothAand B are wrong (b)Ais correct but B iswrong (c) B is correct butAis wrong (d) BothAand B are correct Sol: (c) Line spectra contain information about atoms andmolecules both. C36. Ultraviolet light ofwavelength 800Åand 700Åwhen allowed to fallon hydrogen atoms in their ground state is found to liberate electronswith kinetic energy 1.8 eVand 4.0 eVrespectively. Find the value of the Planck constant. Sol. hv = E0 + T where E0 = ground level energy and T = kinetic energy of electron 0 hc . E . T . . 10 hc 800 .10. = E0 + 1.8 × 1.6 × 10�19 and 10 hc 700 .10. = E0 + 4.0 × 1.6 × 10�19

Subtracting 8 hc 1 1 10. 7 8 . . . .. .. = 2.2 × 1.6 × 10�19 or h = 27 8 2.2 1.6 10 56 3 10 . . . . . = 6.57 × 10�34 Js

ATOMIC PHYSICS www.physicsashok.in 52 C37: The excitation energy of a hydrogenlike ion in its first excited state is 40.8 eV. Find the energy needed to remove the electronfromthe ion. Sol: The excitationenergy in the first excited state is E = 13.6Z2 2 2 1 1 1 2 . . . . . . . 40.8 = (13.6 eV) × Z2 × 34 . Z = 2 Now, ionization energy= 2 2 13.6Z 1 = �4 × (13.6 eV) Eion = �54.4 eV C38: The first ionization potentialof some hydrogen likeBohr atomis xV.Then the value of the first excitation potential for this atomwillbe: (a) xV (b) x2 V (c) 34 xV (d) 20 xV Sol: The value of first excitation potentialis given by x. = x 1 14 . . . . . . . V x. = 3x 4 V Example 34:Adoublyionised lithiumatomis hydrogen likewith atomic number 3. Find thewavelength of the radiation required to excite the electron inLi++ fromthe first to the thirdBohr orbit. (Take ionization energy ofH-atomequal 13.6 eV). Sol: For E1, Z = 3, n = 1 . E1 = � 2 2 13.6Z n = �13.6 9 1. = 122.4 eV E n = 3 3 E n = 2 2 E n = 1 1

.E For E3, Z = 3, n = 3 . E3 = � 2 2 13.6 3 3. = �13.6eV We have .E + E1 = E3 . .E = E3 � E1 = 108.8 eV . . = hc.E = 12400 108.8 Å = 114 Å Example 35: InBohr�smodelof hydrogen atomwhenthe electron ismoving in one of the stationaryorbits then: (a) velocityof the electron is fixed and no emission of energy takes place (b) velocitychanges continuouslybut no emission of energytakes place (c) energyis emitted but the velocitydoes not change (d) energyis emitted and the velocity also changes. �e �e �e �e v1 v2 v3 v4 . Sol: (b) According to the second postulate ofBohr. 1 | v | . = 2 | v | . = 3 | v | . = 4 | v | . = v

ATOMIC PHYSICS www.physicsashok.in 53 Example 36. An electron in an unexcited hydrogenatomacquired an energyof 12.1 eV. Towhat energy level did it jump?How many spectral linesmay be emitted in the course of transition to lower energy levels? Calculate the shortest wavelength. Sol. E = � E0 22 Zn where E0 = 2.18 × 10�18 J = 1 rydberg .E (energygap between unexcited state (n = 1) and an excited state (n= m)) 2 0 2 2 E Z 1 1 1 m . . . . .. .. . 12.1 × 1.6 × 10�19 = 2.18 × 10�18 × 12(1 � 1/m2) n = 3 n = 2 or 1 � n = 1 2 1 m = 0.888 or m= 3 Three lines are emitted. The shortestwavelength corresponds to the greatest energygap. . min hc . = E3 � E1 = E0 2 2 1 1 1 3 . . . .. .. or .min = 34 8 18 6.6 10 3 10 9 2.18 10 8 . . . . . . . . = 1.02 × 10�7 m = 1020 Å Example 37. Ahydrogen like atom(atomic number Z) is in a higher excited state of quantumnumber n.The excited atomcanmake a transition to the first excited state bysuccessivelyemitting two photons of energy 10.2 and 17.0 eVrespectively.Alternatively, the atomfromthe same excited state canmake a transition to the second excited state bysuccessivelyemitting two photons ofenergies 4.25 eVand 5.95 eVrespectively. Determine the values of n and Z. (lonization energyofHatom= 13.6 eV) Sol. Fromthe given conditions En � E2 = (10.2 + 17) eV = 27.2 eV ...(1) and En � E3 = (4.25 + 5.95) eV = 10.2 eV ...(1) Equation (1) and (2) gives E3 � E2 = 17.0 eV or Z2 (13.6)(1/4 � 1/9) = 17.0

. Z2 (13.6) (5/36) = 17.0 . Z2 = 9 . Z = 3 Fromequation (1) Z2 (13.6) (1.4 � 1/n2) = 27.2 or (3)2(13.6) (1/4 � 1/n2) = 27.2 or 1/4 � 1/n2 = 0.222 or 1/n2 = 0.0278 or n2 = 36 . n = 6 ATOMIC EXCITATION WITH THE HELP OF COLLISION (a) An atomcanbe excited to an energyabove its ground state bya collisionwith another particle inwhich part of their joint kinetic energy is absorbed by the atom. (b) An excited atomreturns to its ground state in an average of 10�8 s by emitting one ormore photons.

ATOMIC PHYSICS www.physicsashok.in 54 (c) Energytransfer is amaximumwhenthe colliding particles have the samemass. (d) The energy used in this process will be of discrete nature. (.E = E2 � E1 = hv = hc . ). NOTE: If the joint kinetic energy of colliding particles is less than 20.4 eV (considering particles as Hatoms or one neutron and one H-(atom) then the nature of collision will be necessarily elastic. Example 38: Aneutronmovingwith speed vmakes a head-on collisionwith a hydrogen atomin ground state kept at rest. Find the minimumkinetic energy of the neutron for which inelastic (completely or partially) collisionmay take place. The mass of neutron= mass of hydrogen = 1.67 × 10�27 kg. Sol: Let us suppose that neutron andH-atommove at speeds v1 and v2 after the collision. Suppose an energy .E is used in thisway. On the basis of conservation of linearmomentumand energy,we canwrite mv = mv1 +mv2 .....(i) 12 mv2 = 21 1 mv 2 + 221 mv 2 + .E .....(ii) Fromequation (i) we have v2 = 21 v + 22v + 2v1 v2 .....(iii) Now, fromequation (ii) v2 = 21 v + 22v + 2 E m. .....(iv) . Fromequation(iii) and (iv) 2v1v2 = 2 E m. Hence, (v1 � v2)2 = (v1 + v2)2 � 4v1v2 = v2 � 4 E m. Since v1 � v2 is always real, v2 � 4 E m. . 0 mv2 . 4.E . 12 mv2 . 2.E

Theminimumkinetic energyof the neutron needed for aninelastic collision corresponds to transition from n = 1 to n = 2 . Kmin = 2min 1 mv 2 = 2 × 10.2 eV . Kmin = 20.4 eV Example 39. Consider anexcited hydrogen atominstate nmovingwith a velocityv (v<< c). It emits a photon inthedirectionofitsmotionand changes its state to a lowr statem.Applymomentumandenergyconservation principle to calculate the frequencyv of the emitted radiation. Compare thiswiththe frequencyv0 emitted if the atomwere at rest. Sol. Let En and Em be the energies of electron in nth andmth states. Then

ATOMIC PHYSICS www.physicsashok.in 55 En � Em = hv0 ...(1) In the second casewhen the atomismovingwith a velocity v. Let v´ be the velocityof atomafter emitting the photon.Applyingconservationof linearmomentum, v v´ m m v mv = mv´ + hv c (m=mass of hydrogen atom) or v´ = v hv mc . . . .. .. ...(2) Applying conservationof energy En + 12 mv2 = Em + 12 mv´2 + hv or hv = (En � Em) + 12 m(v2 � v´2) hv = hv0 + 12 m 2 v2 v hmc . . . . . . . . . . . .. . . .. hv = hv0 + 12 m 2 2 2 2 2 2 v v h 2h v m c mc . . . . . . . . . . . hv = hv0 + 2 22 h v h c 2mc . . . Here the termis 2 22 h2mc . is very small. So, can be neglected . hv = hv0 + h v h v

c c . . . or v 1 vc . . . .. .. v0 or v = v0 1 1 vc . . . . .. .. ; v . v0 1 vc . . . .. .. as v < < c Example 40: Thewavelength ofD1 andD2 lines of sodiumare 5890Åand 5896Årespectively, if theirmean wavelength is 6000Åthen find the difference of excited energy states. Sol: E = hc . ....E = 2 hc . .. .E = 34 8 10 20 6.62 10 3 10 6 10 6000 6000 10 . . . . . . . . . . . .E = 3.31 × 10�22 J . .E = 22 19 3.31 10 1.6 10 . . . . , ........2 × 10�3 eV C39:Alithiumatomhas three electrons.Assume the following simple picture of the atom.Two electronsmove

ATOMIC PHYSICS www.physicsashok.in 56 close to the nucleusmaking up a spherical cloud aroundit and the thirdmoves outside this cloud in a circular orbit. Bohr�smodel can be used for themotion of this third electron but n = 1 states are not available to it. Calculate the ionization energyof lithiumin ground state using the above picture. Sol: In this picture, the third electronmoves in the field of a total charge +3e �2e = +e. Thus, the energy are the same as that of hydrogen atoms. The lowest energy is E2 = 1 E4 = 13.6eV 4 . = �3.4 eV Thus , the ionization energy of the atomin this picture is 3.4 eV. Example 41: Find the wavelengths in a hydrogen spectrumbetween the range 500 nmto 700 nm. Sol: The energy of a photon ofwavelength 500 nmis hc . = 1242eV nm 500nm. = 2.44 eV The energy of a photon ofwavelength 700 nmis hc . = 1242eV nm 700nm. = 1.77 eV The energydifference between the states involved in the transition should, therefore, be between 1.77 eV and 2.44 eV. Figure shows some of the energies of hydrogen states. It is clear that onlythose transitionswhich and at n = 2mayemit photons of energybetween 1.77 eVand 2.44 eV. Out of these only n = 3 to n = 2 falls in the proper range. The energyof the photon emitted in the transition n = 3 to n= 2 is E . . = (3.4 � 1.5)eV= 1.9 eV.Thewavelength is . = hc .E = 1242eV nm 1.9eV. = 654 nm. Example 42: Calculate the (a) velocity, (b) energy, and (c) frequency of the electron in first Bohr orbit of hydrogen atom. Sol. (a)We have, vn = 2

0 Ze 2. nh ; but here Z = 1 and n = 1 . v1 = 2 0 e 2. nh = 19 2 9 34 (1.6 10 ) 1 36 10 2 1 6.62 10 . . . . . .. . . . = 2.18 × 106 m/sec (b) We have, En = � 4 2 2 2 2 0 me Z 8. n h Again here, z = 1 and n = 1

ATOMIC PHYSICS www.physicsashok.in 57 . E1 = � 31 19 4 2 9 2 2 34 2 9.1 10 (1.6 10 ) 1 (4 9 10 ) 8 1 (6.62 10 ) . . . . . . . . .. . . . . = �21.758 × 10�19 joule = � 19 19 21.758 10 1.6 10 . . . . = �13.6 eV (c) We have, v = 2 4 2 3 3 0 Z me 4. n h ; here also n = 1 and Z = 1 = 4 2 3 0 me 4. h = 6.57 × 1015 Hz. C40: Find out the radius ofthe hydrogen atomin ground state. Sol. We have, rn = 2 2 0 2 h n mZe. . ; here Z = 1 and n = 1 r1 = 34 2 9 19 2 31 (6.62 10 ) 1 4 9 10 1 (1.6 10 ) (9.1 10 ) . . . . . .. . . .. . . . . r1 = 0.53 Å Example 43: If thewavelength of the firstmember of the Balmer series of hydrogen spectrumis 6562Å, then calculate thewavelength of firstmember ofLymen series in the same spectrum. Sol. We have, for the first member of the Balmer series v1 = R 2 2

1 1 2 3 . . . . . . . = 5 36 R and for the first member ofLyman series, v2 = R 2 2 1 1 1 2 . . . . . . . = 3R4 . 12 .. = 12 .. = 5R 36 × 4 3R = 5 27 . .2 = 1 527 . = 5 6552 27 . = 1215.18 Å Example 44.Thehydrogenatominits groundstate is excitedbymeans ofmonochromatic radiationsofwavelength 975Å.Howmanydifferent lines are possible in the resulting spectrum?Calculate the longest wavgelength among them.Youmayassume the ionizationenergy for hydrogen atomto be 13.6 eV, the Planck constant = 6.63 × 10�34 Js. Sol. En= � E0Z2/n2. The energyrequired to take the electronfromn = 1 to infinityis the ionization energyof the hydrogen atom. . 13.6 = E0(1/12 � 1/.) or E0 = 13.6 eV

ATOMIC PHYSICS www.physicsashok.in 58 Therefore, for hydrogen En = �13.6/n2 eV The energyof the photon incident on hydrogen is 34 8 18 10 hc 6.63 10 3 10 E h 2.04 10 J 975 10 . . . . . . . . . . . . . Let the electron jump fromn= 1 to n=mafter absorbing the incident photon. Then .E = Em � E1 = 13.6(1/12 � 1/m2)eV = 13.6(1 � 1/m2) × 1.6 × 10�19 J . 13.6(1 � 1/m2) × 1.6 × 10�19 = 2.04 × 10�18 . (1 � 1/m2) = 0.9375 or m= 4 The resulting transitions are shown in the figure. So there are six possible lines. The longest wavelength corresponds to theminimumenergygap.Hence longest wavelengthcorresponds to transitionfromm= 4 to m= 3 . 0 2 2 h E 1 1 3 4 . . . . . .. .. . 0 hc E 1 1 9 16 . . . . . .. .. E0 m = 4 m = 3 m = 2 m = 1 . 0 hc 144 7E . . . . 34 8 19 6.63 10 3 10 144 7 13.6 1.6 10 . . . . . . . . . . . . = 1.88 × 10�6 m = 18800 Å. Example 45: The energy of an excited hydrogen atomis �3.4 eV. Calculate the angular momentumof the electron according to Bohr�s theory. Ans: 2.11 × 10�2 joule × sec. Since the energy of an electron in nth level in hydrogen atomis

En = � 2 RCh n or, �3.4 = � 2 13.6 n . RCh = 13.6 eV . n = 2 FromBohr�s theory, L= n h 2. . L = 2 × 0.6 10 34 2 3.14. . . = 2.11 × 10�3 joule sec. REDUCED MASS .In our earlier discussionwe have assumed that the nucleus (a protonin case of hydrogenatom) remains at rest.With this assumption the values of the Rydberg constant R and the ionization energy of hydrogen predicted byBohr�s analysis arewithin 0.1%ofthemeasured values. Rather the proton and electron both revolve in circular orbits about their common centre ofmass.We can

ATOMIC PHYSICS www.physicsashok.in 59 take themotion of the nucleus into account simplybyreplacing themass ofelectronmbythe reducedmass µ of the electron and nucleus. Here . = Mm M.m .....(i) where M= mass of nucleus. The reduced mass can also be written as, . = m1 mM . Now, whenM> > m, mM . 0 or . . m For ordinary hydrogen we let M = 1836.2 m. Substituting in equation (i),we get µ = 0.99946mwhen this value is used instead of the electronmassmin the Bohr equations, the predicted values arewellwithin 0.1%of themeasured values. + � cm m m v v Separation r Applying the Bohr model to positronium. The electron and the positron revolve about their common centre of mass, which is located midway between them because they have equal mass The concept of reduced mass has other applications.Apositron has the same rest mass as an electron but a charge +e. Apositroniumatomconsists of an electron and a positron, each withmassm, in orbit around their common centre ofmass. This structure lasts only about 10�6 s before two particles annihilate (combine) one another and disappear, but this is enough time to studythe positroniumspectrum. The reducedmass ism/2, so the energy levels and photon frequencies have exactlyhalf the values for the simpleBohrmodelwithinfinite protonmass. CM M m r2 r1 Now, let us provewhymis replaced bythe reducedmass .when motion of nucleus (proton) is also to be considered. In figure both the nucleus (mass =M, charge = e) and electron (mass =m, charge =e ) revolve about their centre ofmass (CM) with same angular velocity (.) but different linear speeds. Let r1 and r2 be the distance ofCMfromproton and electron. Let r be the distance between the proton and the electron. Then, Mr1 = mr2 .....(ii) r1 + r2 = r .....(iii) . r1 = mr M.m and r2 = Mr M.m .....(iv) Centripetal force to the electron is provided by the electrostatic force. So, mr2.2 =

22 0 1 e 4.. r or m Mr M r . . . . . . . .2 = 22 0 1 . e 4.. r or Mm M m . . . . . . . r3.2 = 2 0 e 4..

ATOMIC PHYSICS www.physicsashok.in 60 or .r3.2 = 2 0 e 4.. .....(v) where Mm M.m = . Moment of inertia of atomabout CM, I = 2 2 1 2 Mr .mr = Mm M M . . . . . . . r2 = .r2 .....(vi) According toBohr�s theory, nh 2. = I... µr2..= nh 2. .....(vii) Solving equations (v) and (vii) for r,we get r = 2 2 0 2 n h e .. . .....(viii) Electrical potential energyof the system, U = 20 e 4 r ... and kinetic energy,K = 12 I.2 = 12 µr2.2 Fromequation (v),.2 = 2 3 0 e 4.. .r , K = 20 e 8.. r .Total energyof the system, E = K + U = � 20 e 8.. r

Substituting value of r fromequation(viii),we have E = � 4 2 2 2 0 e 8 n h . . .....(ix) The expression for Enwithout considering themotion of proton is En = � 4 2 2 2 0me 8. n h , i.e.,mis replaced byµ while considering themotionof proton. NOTE :(i) Variation of rn, vn and Enwithmass of election is as under, rn . 1m , vn = independent ofmand En .m Sometimes the electron is replaced bysome another particlewhich has a charge �e but mass different fromthemass of electron. Here, two cases are possible. Case 1: Let saymass of the replaced particle is x times themass of the electron and nucleus is still very heavycompared to the replaced particle, i.e., themotion of the nucleus is not to be considered. Inthis case

ATOMIC PHYSICS www.physicsashok.in 61 rnwill become 1x times, vn will remain unchanged and En becomes x times. Case 2: In this case motion of nucleus is also to be considered, i.e., mass of the replaced particle is comparable to themass of the nucleus. In this case themass of the electron is replaced bythe reducedmass of the nucleus and the replaced particle. Let say the reducedmass is ytime themass of the electron. Then, rn will become 1y times, vn remains unchanged and En becomes y-times. (ii)Reduced massm= 1 2 1 2 m m m . m ofm1 and m2 is less than both the masses. C41:Apositroniumatomis a systemthat consist of a positron and an electron that orbit each other. Compare thewavelengths of the spectral lines of positroniumwith those of ordinaryhydrogen. Sol: In this case reducedmasswillbe given by m. = mM m.M = m2 2m = m2 wherem=mass of the electron. Hence, the energy levels of a positroniumatomare E.n = m' m . . . . . . 12 En = 12 E 2n It means that the Rydberg constant for positroniumis half as large as it is for H-atom. As a result the wavelength inthe positroniumspectrallines are all twice thoseof the corresponding lines in theH-spectrum. Example 46: Bohr�s theoryassumes that nucleus is of infinitemass and so electron rotates round the stationary nucleus. Assuming the nucleus to be of finite mass MH, the value of correct Rydberg constant will be (consider hydrogen atom) (a) 4 H 2 3

H 0 mM e M m 8 ch . . . . . . . . (b) 4 H 2 2 3 H 0 M m e mM 8 c h . . . . . . . . (c) 4 H 2 3 H 0 mM e M m 8 ch . . . . . . . . (d) 4 2 3 0e 8. ch Sol: (c) In this case both electronand nucleuswill rotate about a common centre ofmass, sayO. Suppose that the radiiof the electron and nucleus orbits are re and rn respectivelythenbydefinition of centre ofmass; the electron, the centre ofmassOand the nucleus are always in a straight line. . MHrn =mre whenmis themass of electron . n e n r r . r = H m m.M Let r = re + rn then rn = H mr m.M and re = H H M r m.M FromBohr�s quantizationrule,we have

ATOMIC PHYSICS www.physicsashok.in 62 2 2 m.re .MH.rn = nh 2. [.is the angular velocity about the centre ofmass] Putting thevaluesof rn and re,we get HH mM m M . . . . . . . .r2 = nh 2. The above equation can be comparedwithm.r2 = nh 2. . TheRydberg constantwill be given by R = HH mM m M . . . . . . . 4 2 3 0e 8. ch Obviously the reducedmass of the electron is HH mM m M . . . . . . . . Example 47.Apositromiumatomis a bound systemofan electron (e�) and its antiparticle positron (e+) revolving about their centre ofmass. Find thewavelength of the radiationwhen the systemde-excites fromits first excited state to the ground state. Sol. This problemcan be solved byBohr�s theoryof the hydrogen atombyreplacing themass of the electron by its reducedmass. . . e e e e e m m m µ m m 2 . . . . . 2 4 e n 2 2 2 0

Z e mE 2 8 h n . . . .. .. . .4 2 e n 2 2 2 0 e m Z E 8 h 2n . . . . . .. . . . . . . . . . En = � 2.18 × 10�18(12/2n2) (. e4me/8.02h2 = 2.18 × 10�18) When n = 1, E1 = � 2.18 × 10�18/2 When n = 2, E1 = � 2.18 × 10�18/8 . .E = E1 � E2 = 2.18 × 10�18(1/2 � 1/8) = 2.18 × 10�18 × 3/8 . .E = hc/. = 2.18 × 10�18 × 0.375 . . . . . 34 8 18 6.63 10 3 10 2.18 10 0.375 .. . . . . . . . , . = 1.2165 × 10�7 m = 2433 Å Example 48:Themass ofmuon (. ) is 207 times that of the electron and charge = �1.6 × 10�19 C.Amuon can be captured by a nucleus to formamuonic atom. Calculate the value of ionization energyof themuonic atom.

ATOMIC PHYSICS www.physicsashok.in 63 Sol: The ionization energyof themuonic atomis obtained by replacingme inH-atomformula bythe reduced massmof the proton-muon system. This reducedmass is m = p p m m m m. . . = e e e e 1836m 207m 1836m 207m .. . 186me Thus the ground state energy is (n = 1, Z= 1). 2 2 4 1 2 2 mk e E h . . . 2 2 4 e 1 2 e m 2 m k e E m h . . . . . . . . . E1 = � 186 × 13.6 eV The ionizationenergy = �E1 = 186 × 13.6 eV = 2.53 KeV PRODUCTION OF X-RAYS X-rayswere accidentallydiscovered byWilhelmRontgenin 1895 during the course of some experiments with a discharge tube.At present, it is well known that these rays are produced whenever fast moving electrons strike a high atomicweight solid like tungsten kept in vacuum. (a) X-ray Tube: The essentialelements of amodernCoolidgeX-rayvacuumtubewhich iswidely used for commercial and medical purposes are shown in figure. Electrons are produced thermionically froma tungsten filamentary cathode F which is heated to incandescence either by a storage battery or by a low-voltage alternating current froma stepdown transformer T2. These electrons are focussed on the target T with the help of a cylindrical shield Swhichsurrounds Fand ismaintained at a negative potential. The electrons are accelerated to veryhigh speeds (upto 10%of velocity of light) by the d.c. potentialdifference (of about 50 kV- 100 kV) applied between F and the anode (also called anticathode). This high d.c. potential is obtained froma step-up transformer T1whose output is converted into direct current by full-wave

rectifier and a suitable filter. � + 50 kV ElectronsT A Cooling Tube Fins X-rays R F S B The targetTusuallyemployed inX-raytubes is amassive block oftungstenor inmanycases, amolybdenum

ATOMIC PHYSICS ATOMIC PHYSICS plug embedded in the face of a solid copper anode. The face of the copper anode is sloped at about 45º to the electron beam. Being very good conductor of heat, copper helps to conduct heat efficiently to the externalcooling finsorthewater-coolingsystem.Undertheterrificbombardmentofthetargetbysomany electrons,mostmetalswillmelt. Thatiswhymetalsliketungsten,platinumandmolybdenumetc.areused whichhave highmelting points and also have a highatomicweight (whichis essentialfor abundant production of X-rays).

When the electrons strike the tungsten target, they give up their kinetic energyand thereby produce X-rays. T A Cooling Water Tube X-rays R F ST1 T2 Rectifier (b) Control of Intensity and Quality The intensity of X-rays depends on the number of electrons striking the target. This number is determined by the temperature of the electron-emitting filament which itself is proportionalto the heater current. Hence by controlling the filament current with the help of a rheostat R, thermionic emission and hence intensity of X-rays can be controlled. The quality of X-rays is measured in terms of their penetrating power which is dependent on the potential difference between filamentary cathode and the anode. Greater this accelerating voltage, higher the speed of the striking electrons and consequently, more penetrating the X-rays produced. It is customaryto refer to highly penetrating X-rays (i.e. those possessing high frequency) as hard X-rays and to those less penetrating (i.e. of low frequency) as soft X-rays. Obviously, the quality or penetrating power of X-rayscan be controlled by varying the potential difference between the cathode and anode. It will be noticed from above explanation that in coolidge X-ray tube, it is possible to achieve separate control ofthe intensity and quality of X-rays independent of each other. It has beenfound that apart fromthe intensity and quality ofX-rays, their abundance depends on the atomic weight ofthe target material. Target materials of higher atomic weights yield a greater abundance of X-rays than those oflower atomic weights.

ORIGIN OF X-RAYS X-rays are produced when high-speed electrons strike some material object. However, majority of the electronsthatstrikeasolidtarget,donothingspectacular atall.Most ofthemundergoglancingcollisions

with the matter particles, lose their energy a little bit at a time and thus merely increase the average kinetic energyofthe particles of the target material. It is found that nearly99.8 percent of the energyof the electron beam goes into heating the target. But a smallnumber ofthe bombardingelectrons produce X-rays bylosing their kinetic energyinthe following two ways:

(i) Someofthehigh-velocityelectronspenetratetheinterioroftheatomsofthetargetmaterialandareattracted www.physicsashok.in 64

ATOMIC PHYSICS www.physicsashok.in 65 bythe positive charge of their nuclei.As an electronpasses close to the positive nucleus, it is deflected from its pathas shownin figure. The electronexperiences deacceleration duringits deflectionin the strong field of the nucleus. The energy lost during this de-acceleration is given off in the formofX-rays of continuously varyingwavelength(and hence frequency). TheseX-rays produce continuous spectrumwhen analysed by Bragg spectrometer.This spectrumhas a sharplydefined short-wavelength limit .min(or high-frequencylimit fmax)which corresponds to themaximumenergyofthe incident electron. X-RayX-Ray + + + + X-Ray 1 v. 2 mv.2 Continuous spectrum12 mv2v If, as shownin figure, the striking electronhas its velocityreduced fromvto v. during its passage through the atomofthe targetmaterial, then its loss of energyis = ( 12 mv2 � 12 mv.2). Thismust equal the energyof the X-rayphotons emitted. . 12 m(v2 � v.2) = hv The highest ormaximumfrequency of the emitted X-rays corresponds to the case when the electron is completely stopped i.e. when v. = 0. In that case 12 mv2 = hvmax .....(i) If the electron is accelerated through a potential ofVvolts, then 12 mv2 = eV .....(ii) From(i) and (ii), we get hvmax= eV; vmax = eV/h Now, hvmax= hc/.min [. c = v ...] . hc/.min = eV or .min = hc/eV Substituting the values of e = 1.602 × 10�19C h = 6.62 × 10�34 J-s and c = 3 × 108 m/s, we get .min = 34 8 19 6.62 10 3 10 1.602 10 V . . . . .

. = 1.24 10 6 V . . m or .min = 12400 V Å [1Å = 10�10 m] SuchX-rays are veryaptly called �braking� radiations because they are due to braking or slowing down of high-velocity electrons is the positive field of a nucleus. These radiations constitute, as said earlier, the continuous spectrumof theX-rays because they consist of a series of uninterruptedwavelengths having a sharplydefined short-wavelength limit .min. TheseX-rays are independent ofthe nature of the targetmaterial but are determined by the potentialdifference between the cathode and anode of theX-ray tube. (ii) Some ofthe high-velocityelectronswhile penetrating the interior ofthe atoms of the targetmaterial, knock

ATOMIC PHYSICS www.physicsashok.in 66 off the tightlybound electrons in the innermost shells (likeK, L-shells etc) of the atoms.When electrons fromouter orbits jump to fillup the vacancyso produce, the energydifference is given out in the formofXrays of definite wavelength (and frequency). These wavelengths constitute the line spectrumwhich is characteristic of thematerial of the target. KL M e e e (a) KL M (b) X-Ray K -line . Figure (a) shows the case when the high-velocity incident electron knocks off one electron fromtheKshell.As shown in figure (b), this vacancyinK-shellis filled by a nearby electron in the L-shell. During the jump anX-rayradiation is emittedwhose frequencyis given by Ek � E. = hv where Ek is the energyrequired to dislodge an electron formtheK-shell and El is that required for L-shell. Since this energy difference is comparativelyvery large, theX-rays emitted have verylarge energycontent and hence are highly penetrating. If, however, this vacancy inK-shell is filled up by an electron jumping fromM-shell, theX-rays emitted would be stillmore energetic andwould consequentlypossess stillhigher frequencybecause .E =(Ek �Em) ismore than .E = (Ek � EL). SuchX-rays arising frommillions of atoms produce the K-lines as shown in figure. Usually, two linesK. andK. of this series are detected although there aremanymore. Similarly, when the incident electron carries somewhat lesser amount of energy, it dislodges an electron fromthe L-orbit and an electron either fromM-orbit or other outer orbits takes its place so that X-rays of frequency lower than that of theK-series are produced. This gives the L-series of theX-ray spectrumas shown byK., L. and L. lines in figure (a). KLMN O K. K. L. L. L. M (a) E = 0 �2 �20

�200 �2000 �20,000 K. K.L. L. L. M. M. KLMNO .SHELL Energy in eV (b) M Spectral lines ofM-series are produced ina similarway as shownin the energy-leveldiagramof figure (b). As stated earlier, theseK, LandMseries constitute the line spectra of theX-rayswhich are characteristic of thematerialused as target in theX-ray tube. Hence, theX-rays produced byanX-ray tube consist of two parts: (i) one part consists ofa series of uninterruptedwavelengths having a short cut-offwavelength .min.This

ATOMIC PHYSICS www.physicsashok.in 67 constitute the continuous spectrumand (ii) the other part consists of a number of distinct and discrets wavelengths which constitute the line or discontinues spectrumof theX-rays. X-ray Spectrum As explained inX-ray spectrumconsists of (i) continues spectrumand (ii) line spectrum. These two are shownin figure. (a) Continuous Spectrum (i)It is produced due to the de-acceleration of high-velocityelectronswhen theyare deflectedwhile passing near the positively, charged nucleus of an atomof the targetmaterial. K. K. L. L. L. Intensity Continuous Spectrum .min . f Continuous Spectrum K. K. L. L. L. (a) (ii) It has a sharply-defined short wavelength limit given by .min = 12.400 V × 10�10 m or 12, 400 V Å (iii)The cut-offwavelength.min is independent ofthenature ofthe targetmaterialbut is inverselyproportional to the potentialdifference between the cathode and anode of anX-ray tube. The value of .min decreases as this potentialdifference is increased. (iv) The intensity of the continuous spectrum(given by the area enclosed bythe curve of figure (b) is found verynearlyproportional to the square of the applied voltage for a given target and to the atomic number of the target materialwhen a constant potential difference is applied. 0 1.0 2.0 3.0 4.0 5 kV 10 kV 15 kV 20 kV K 25 kV . K. L. L. L. Wave length(A.U.) X-ray Intensity Tungsten Target (b) (v) There is a shift of themaximumintensity position towards the short wavelength side as voltage is increased. (b) Line Spectrum

(i) It is producedwhen electrons are dislodged fromthe innermost orbits ofthe atoms ofthe targetmaterialfollowed byelectronjumps fromouter orbits. (ii) It consists of discrete spectral lineswhichconstituteK-series, L-series andM-series etc.K-series consists of those lines for which electron jumps end at K-level. (iii) K-series beingmost energetic constitute the hardX-rayswhereasL- andM-series formthe �soft�Xrays. (iv) Line spectrumis characteristic of the targetmaterial used. In

ATOMIC PHYSICS www.physicsashok.in 68 fact, X-rays constituting the line spectrum are known as characteristicX-rays. The number of lines present in the spectrum depends both on the nature of target material and the excitation voltage. K. K. 90 79 50 42 28 Cr 24 Cu Mo Sn Au Cf . Mass No. (Z) (c) (v) There is a regular shift towards shorter wavelength in theKspectrumas the atomic number ofthe target is increased figure (c). The exact relationship, as found byMoseley, is 21 .. = 12 .. = 2 1 2 2 (Z 1) (Z 1) .. where v1 is the frequency of theK. line for a target material having an atomic number ofZ1 and v2 and Z2 are similar quantities for some different targetmaterial. C42:AnX-raytubeworks on 60,000V.What will be the wavelength ofX-rayemitted in it. Sol: .min = 12, 400 V Å Here, V = 60,000 V . .min= 12, 400 60,000 = 0.2 Å C43: If the potential difference applied across anX-ray tube is 12.4 kV and the current through it is 2 mA, calculate: (i) the number of electrons striking the target per second (ii) the speedwithwhich theystrike it (iii) the shortest wavelength emitted Take e = 1.6 × 10�19 C and m = 9.1 × 10�31 kg

Sol: (i) If n is the number of electrons striking the anode per second, then I = ne . n = Ie = 319 2 10 1.6 10.. .. = 1.25 × 1016 s�1 (ii). v 2eV 5.93 105 V m . . . v = 5.93 × 105 v = 5.93 × 105 12, 400 = 6.6 × 107 m/s (iii) .min = 12, 400 V = 12400 12400 = 1Å C44: Calculate theminimumapplied potentialrequired to produceX-rays of 1Åwavelength. Sol: .min = 12400 V Å . V = min 12400 . = 12400 1 = 12.4 kV C45:An X-ray tube passes 5 mAat a potential difference of 100 kV. Calculate the maximumspeed of the

ATOMIC PHYSICS www.physicsashok.in 69 electrons striking the target and the rate of productionofheat at the target if only0.1 percent of the incident energy is converted into X-radiations. Take e/m= 1.76 × 1011 C/kg and J = 4.18 joules/cal. Sol: . v 2eV 5.93 105 V m . . . v = 5.93 × 105 100,000 = 1.88 × 108 m/s � + 100 kv 5 mA Incident power = 100,000 × 5 × 10�3 = 500W Power converted into heat = 99.9%of 500 = 499.5W Heat produced / second = 499.5 4.18 = 119 cal/s. C46:AnX-raytube operated at 30 kVemits a continuousX-rayspectrumwith a short wavelength limit .min = 0.414 Å. Calculate Planck�s constant h if e = 1.602 × 10�19 C and c = 3 × 108m/s. Sol: .min = ch eV . h = min eVc. = 19 3 19 8 1.602 10 30 10 0.414 10 3 10 . . . . . . . . = 6.63 × 10�34 J-s MOSELEY�S LAW In 1913-14,Moseleycarried out a systematic study of the characteristicX-ray spectra ofvarious elements used as targets in anX-raytube.By usingBragg�s spectrometer for the purpose, the remarkably similar to each other in the sense that each consists of K-L and M-series. However, there is one very important difference. The frequency of lines (in every series) produced froman element of higher atomic number is greater thanthat produced byanelement of lower atomic number. It is due to the fact that binding energyof electrons increases aswe go fromone element to another ofhigher atomic number. Because there is greater positive charge onthe nucleus ofanelement of higher atomic number, larger amount ofenergyis required to liberte an electron fromtheK, L andMshells of that element. Consider theK. line of the characteristicX-rayspectrumof any element. It is found that higher the atomic number ofthe targetmaterial, higher is the frequencyof theK. line produced byit. The exactmathematical relationship betweenfrequency and atomic number is given by v . (Z � b)2 or . . (Z � b) or . = a(Z � b)

where Z is the atomic number of the element and a and b are constants for a particular series but varyfrom one series to another i.e. their values for K-series are different fromthose for L-series etc. The constant b is known as nuclear screening constant. For lines ofK-series, b = 1. Its values for lines ofL-series ismore. The above relation is known asMoseleylawfor the characteristic or lineX-ray spectrum. Itmaybe stated as follows: The frequency of a spectral line in the characteristic X-ray spectrumvaries directly as the square of the atomic number ofthe element emitting it. Figure showsMoseleydiagramfor K. andK. lineswhich is

ATOMIC PHYSICS www.physicsashok.in 70 obtained by plotting . versus atomic number of different elements of the periodic table.As expected, the graph is linear. 5 10 15 20 25 × 108 Al-13 10 Ca-20 Zn-30 Ze-40 Sn-50 K. K. Atomic number Frequency An exact formofMoseley�s lawis 1. = R(Z � .)2 2 2 1 2 1 1 n n . . . . . . . whereR is Rydberg�s constant, Z the atomic number, . a correction factor and n1 and n2 the principal quantum numbers of the energy levels between which the transition occurs. Importance of Moseley law The great significance ofMoseley law lies in the fact that it proves for the first time that it is the atomic number and not the atomicweight of an elementwhichdetermines its characteristic properties (bothphysical and chemical). It provides the proper guideline that elementsmust be arranged intheperiodic table according to their atomic numbers and not their atomicweights. Accordingly,Moseleylawhas been used to place elements in their proper sequence in the periodic table in certain questionable cases. For example, ifwe go by the atomic weight, potassium(19K39) should come before argon (18A40) and similarly, nickel (28Ni58.7) should precede cobalt (27Co58.9). ButMoselylawdictates that as per their atomic numbers, their order should be just opposite of the above. This fact is further supported bythe chemical properties of these elements. Moseleylawhas led to thediscoveryof newelements like hafnium(72), promethium(61), technetium(43) andrhenium(75) etc. bytheindicationofgaps inMoselydiagram.This lawhas beenalso helpfulindetermining the atomic number of rare earths therebyfixing their position in the periodic table. It can be shown thatMoseleylawis in accordancewithBohr�s theory of spectral emissionfromatoms.As shown inwhenan electron jumps froman orbit n2 to the orbit n1, the frequencyof the radiationgiven out is, v = 4 2 3 0

me 8. h .Z2 2 2 1 2 1 1 n n . . . . . . . Thismaybe put asv = 4 2 3 2 2 0 1 2 me 1 1 8 h n n .. .. .. .. .. . .. .. . .. .. Z2 or v . Z2 or . . Z Bohr did not take into account the screening effect of electrons whereasMoseley did. That is why the expression becomes . . (Z � b) Formulae for K- and L-series of X-ray Spectrum The frequencies of the various lines in the K- and L-series of the X-ray line spectrumare given by the following empiricalformulae. K-series: The general formula is 1. = R(Z � 1)2 2 1 1 n . . . . . . . where n = 2, 3 etc Here, the nuclear screening constant is unity.

ATOMIC PHYSICS www.physicsashok.in 71 (i) For K. line, n = 2. . 1. = R(Z � 1)2 1 14 . . . .. .. = 3R4 (Z � 1)2 (ii)For K. line, n = 3 . 1. = R(Z � 1)2 1 19 . . . .. .. = 8R9 (Z � 1)2 L-Series: The general formula is 1. = R(Z � 7.4)2 2 1 1 4 n . . . . . . . where n = 3, 4 etc. Here, screening constant is 7.4. (i) For H. line, n = 3 . 1. = R(Z � 7.4)2 1 1 4 9 . . . . . . . = 5R 36 (Z � 7.4)2 (ii)For H. line, n = 4 1. = R(Z � 7.4)2 1 1 4 16 . . . . . . . = 3R 16 (Z � 7.4)2 C47: Find the nuclear screening constant for the L-series ofX-rays if it is known that X-rayswith awavelength of . = 1.43Åare emittedwhen an electron in a tungsten atom(Z = 74) is transferred fromtheM-level to L-level. Take Rydberg constant = 10.97 × 106m�1. Sol:When electron jumps fromM to L-level, the first member of the L-series i.e. L. line is given out. Its wavelength as given byMoseley�s lawis 1. = R(Z � b)2 2 2 1 1 2 3

. . . . .

. . = 5R 36 (Z � b)2 Substituting the givenvalues,we have 1010 1.43 = 10.97 × 106 × 5 36 (74 � b)2 . (74 � b)2 = 4589.8; (74 � b) = 67.75 ; b = 6.25. Example 49: The K. and L. absorption edges of copper occur at wavelengths 1.380 Å and 11.288 Å respectively. Calculate the atomic number of copper. Sol: It shouldbe remembered that absorptionages are found inthe absorptionspectrumofX-rays.The absorption edge ofeachseries (ofline spectrum) represents the limit ofthat series. Inotherwords, the short-wavelength limit of each series is called its absorption edge and iswritten as ... It maybe obtained by putting n= . in the formulae.Moreover, forK-series, the value of screening constant forK.member is 3.3 (instead of1 for othermembers). Corresponding value for L-series is 11(instead of 7.4). K-Series: 1. . = R(Z � 3.3)2 2 .1. 1 . . . . . . = R(Z � 3.3)2 . 1010 1.38 = (Z � 3.3)2 L-Series: 1. . = R(Z � 11)2 2 1 1 4 . . . . . . . . = R4 (Z � 11)2

ATOMIC PHYSICS www.physicsashok.in 72 . 1010 11.288 = R4 (Z � 11)2 Dividing one bythe other, we get 2 2 (Z 3.3) (Z 11) .. = 11.288 4.1.38 = 2.045 or (Z 3.3) (Z 11) .. = 1.43 or Z = 29 Example 50:An impure tungsten target emits a strongK. line of . = 0.21Åand aweakK. line of . = 1.537 Å. Can you identify the impuritytaking the nuclear screening constant as unity. Given for tungsten,Z= 74, for Ni = 28. Sol: Thewavelength for K. line is given bythe relation 1. = R(Z � 1)2 1 14 . . . . . . . = 34 R(Z � 1)2 For tungsten 10 1 0.21.10. = 34 R(74 � 1)2 = 15,987R 4 For impurity 10 1 1.537.10. = 34 R(Z � 1)2 Dividing one bythe other, we get 1.537 0.21 = 2 5329 (Z.1) ; Z = 28 Obviously, impurity is nickelwhose atomicmass number is 28. Absorption of X-rays When a narrowandmonochromatic beamofX-rays passes throughmatter, part of it is a

bsorbed and the remaining part is transmitted.Absorption of x-rays can be studiedwith the help of the apparatus shown in figure. TheX-rays produced by anX-raytube are first made into awell defined narrowbeamby passing themthrough two fine slits S1 and S2 in the two lead plates. The beamis thenmonochromatised byBragg reflectionfroma crystal(not showninthe figure) and allowedto enter the ionizationchamberwhichmeasures the ionization current.The strength of the ionization current is ameasure of the intensityof theX-ray. Next, a sheet of the absorbing material is interposed in the path of the X-ray beambefore it enters the ionization chamber. It is found that the ionization current and hence the intensity ofX-rays is reduced by their passage through the absorber sheet. The ratio 0 I I of the two ionization currents can be used to measure the absorptioncoefficient of thematerial.

ATOMIC PHYSICS www.physicsashok.in 73 L1 L2 S1 S2 Absorber Ionization Chamber Let I0 be the initial intensityof a homogeneousX-raybeamincident normallyon an absorber sheet and I the intensity after the beamhas travelled a thickness x of the absorber. If dI is the further decrease in intensityover a thickness dx of the absorber figure, then dI dx gives the rate of decrease of intensity with thickness. Assuming that this rate is proportionalto the intensityI,we have I0 I (I � dI) x dx � dI dx . I or dI dx = �.I where . is a constant of proportionality and is called the linear absorption coefficient of the absorber (it is also known asmacroscopic absorption coefficient or linear attenuation coefficient). Now, dI I = �..dx .....(i) Integrating both sides of the above equation, we get . dI / I = �. . dx . logeI = �.x + K .....(ii) The value ofthe integration constant Kcan be found fromthe known initial conditionswhichare that when z = 0, I = I0. Substituting these values in equation (ii) above, we have I0 x I O logeI0 = K Hence, equation (ii) becomes logeI = �.x + logeI0 or e 0 log I I = �.x or 0 I I = e�.x or I = I0e�.x .....(iii) It is seen that intensity of the X-ray beamdecreases exponentiallywith the thickness of the absorbing material as shown in figure. BRAGG�S LAW . .

A P S Q E F M B G H N R Figure gives a 3-dimensional view of how a beam of monochromaticX-rays undergoesBragg�s reflection from different planes in a NaCl crystal. Figure gives a

ATOMIC PHYSICS www.physicsashok.in 74 2-dimensional viewof the same diagram. It shows a beam ofmonochromaticX-rays incident at a glancing angle . on a set of parallelplanes ofNaCl crystal. The beamis partically reflected at the successive layers rich in atoms Ray no.1 is reflected fromatomAinplane1whereas rayno. 2 is reflected from atom B lying plane 2 immediately below atom A. Whether two reflected rays will be in phase or antiphase with each other will depend on their path difference. This pathdifference can be found bydrawing perpendicularsAMandANon rayNo. 2. Since the two rays travel the same distance frompointsAandNonwards, it is obvious that ray no. 2 travels an extra distance = MB + BN Hence, the path difference between the two reflected beams is = MB + BN M. . N . A . 1 2 3 1 2dd Plane 1 Plane 2 Plane 3 B = d sin. + d sin. = 2d sin. where d is the interplanar spacing i.e. vertical distance between two adjacent planes belonging to the same set. The two reflected beamswillbe in phasewith eachother if this pathdifference equals an integralmultiple of la dnwill be antiphase if it equals an oddmultiple of ./2. Hence, the conditionfor producingmaxima becomes 2d sin. = n. where n= 1, 2, 3 etc., for the first order, second order and third ordermaxima respectively.This equation is known as Bragg�s Law. PROPERTIES OF X-RAYS Main properties ofX-raysmay be summarised as under: (i) Likevisible light,X-rays consist ofelectromagneticwavesofveryshortwavelength(or ofveryhighfrequency) and showreflection, refraction, interference, diffraction and polarisation etc. (ii) They are not deflected by electric andmagnetic field. (iii) They posses high penetrating power and can pass throughmanysolidswhich are opaque to visible light. The transparency depends on the density of the material. Higher the density of the substance, the less transparent it is to theX-rays. For example, sheet of lead 1 cmthick can absorbX-rayswhereas aluminium sheet ofsame thickness cannot. The penetrating power ofX-rays depends upon (a) the voltage applied across the cathode and anode of theX-raytube and (b) the atomic number of thematerialof the cathode.Greater the accelerating potential of the X-ray tube and higher the atomic number of its target material, the more penetrating the X-rays

produced. (iv) They ionize a gas and also eject electrons frommetals onwhich they fall. (v) Theycause fluorescence inmanysubstances like barium, cadmium, tungstate and zinc sulphide etc. (vi) Theysuffer compton scattering. (vii) They have a destructive effect on living tissue. Exposure of humanbodyofX-rays causes the reddening of skin and surface sores. Practical Applications of X-rays On account of their diverse and distinctive properties,X-rays have beenput tomanyuses in different fields of our dailylife i.e. in industry,medicine and research. (a) Industrial applications: Some of these applications are as under:

ATOMIC PHYSICS ATOMIC PHYSICS (i) to detect and photograph defects within a body i.e. in its internal structure such as metals, machine parts intended for withstanding highpressures, cracks in wood, porcelainand other insulators, defectsindiamonds and other precious stones, in moulds, forgings and castings etc. (ii) toanalysesthestructureofalloysandotherrcompositebodiesbydeterminingthecrystalforminaningot with the help of diffraction of X-rays. In this way, alloys like cobalt-nickel, steel, bronze, duraluminium, artificial pearls and old paintings have been analysed. (iii) to study the structure of rubber, cellulose and plastics. The diffraction of X-rays bythese substances leads to valuable information about their molecular grouping. (b) Applications for pure scientific research: (i) for investigating the structure of the atom. (ii) for studying the structure of the crytalline solids and alloys (X-ray crystallography). (iii) for identification ofchemical elements including determination oftheir atomic numbers. (iv) for analysing the structure of coraplex organic molecules by examining their X-ray diffraction patterns. (c) Medical Applications: These can be broadly divided into two classes; one for diagnosis purposes (radiography) and the other for curative purposes (X-ray therapy). (i) Radiography: X-rays are being widely used for detecting fractures, tumours, the presence of foreign matter like bullets etc. in the human body as well as diseased organs of the boy. It is due to differential absorption of X-rays between bones, tissues and metals. Radiographs or X-ray photos are used for this purpose. Since bones are more dense and hence more opaque to X-rays then flesh, a contrasting radiograph ofhuman body can be obtained for leisurely study by interposing it between the X-ray tube and photo film. Where the organs do not provide contrast as, for example, the intestines or other fleshy parts of the human body, a artificial means are adopted to create sufficient contrast between them. In such cases, before taking radiograph, barium or bismuth meal is given to the patient. This meal consists of milk to which some amount of barium sulphate or bismuth carbonate has been added.Afew hours after the meal has been taken, these powders settle in the gastrointestinal tract and if a radiograph is taken at that item, the intestines stand out in sharp contrast to the surrounding tissues due to the fact that absorptioncoefficient ofbarium is greater on account ofitshigh atomic number. In this way, pepticulcers and ruptures etc. in the internal organs ofthe human body can be accuratelylocated. Similarly, radiographs are routinely used for the diagnosis of tuberculosis, stones in kidneys and gall-bladders etc. (ii) X-ray or Rontgen Therapy Many types of skin diseases, malignant sores, cancers and tumours have been cured bycontrolled exposure to X-rays of suitable quality. This curative power of X-rays is due to the fortu

nate fact that diseased tissue is more susceptible to destructionthan the surrounding healthytissue. Unnecessarylong exposure of human body to X-rays produces many injurious effects including the loss of white cells in the blood, sterility and harmfulgenetic changes. X-rays have been used for the identification of different types of cells and tissues and for bringing about genetic mutations.

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MODERN PHYSICS www.physicsashok.in 123 THINKING PROBLEMS X-rays 1. What leads you do believe that X-rays are electromagneticwaves ? 2. The occurrence ofa lower bound ofwavelengths ofX-rays produced inanX-raytube lends support to the quantumconcept ofradiation. Explain how. 3. IfYoung�s experiment is repeatedwith electron beams interference is observed. Does thismean that an electron gets divided into twowhile passing through the slits ? 4. Water irradiatedwithX-rays is unsafe for drinking. Is this true or false ? 5. An electronmoves through a gas-filled region in the presence of a transversemagnetic field. Describe its motion. 6. Fluerescence is produced by ultraviolet rays but never by infrared rays. Explainwhy. 7. Aneutron, a proton, an electron and an alpha particle enter a region of constant magnetic fieldwith equal velocities. Themagnetic field is along the inward normal to the place of the paper.The tracks of the particles are labelled in the figure. Which tracks do the electron and alpha particle follow? 8. Achargedandanunchargedparticlehavethe samemomentum.Willtheyhave the samedeBrogliewavelength? 9. The electricalconductivityofa gas increaseswhenX-rays or .-rays pass throughit.Explainthis phenomenon. 10. Why are tungstenor platinumwidely used as the target theX-raytubes ? 11. Auniformelectric field acts normally on amoving charge. Can the charge be deflected through 90º ? 12. Auniformelectric field acts normally on amoving charge. Iswork done by the field on the charge (a) as it enters the field, (b) later ? 13. Does the speed of a charged particle changewhen (a) a magnetic field, (b) an electric field, acts on it for some time ? 14. Parallel electric and magnetic fields act on on a charged particle moving perpendicular to these fields. Describe its subsequentmotion. 15. Howdo you conclude that cathode rays are fast moving negativelycharged particles ? 16. �Magneticmirror�is a termfor the region of amagnetic field inwhich there is an intense concentration oflines ofinductionas shown inthe figure. Suppose a charged particle approaches amagneticmirror.What will happen to it ? 17. Howis amonoenergetic, slightlydiverging beamof charged particles focussed byamagnetic field ? 18. X-rays are producedwhen a fast electron hits a proper target.What happens to the electron ? 19. Why does the target in anX-ray tube become hot ? 20. X-rays can be produced in cathode ray tubes and also in Coolidge tubes.Why are the latter preferred in actual use ? 21. Why is thewave nature ofmatter not apparent in our daily lives ? 22. Aneutralpion decays into two gamma photons. .0 .... Why cannot a single photon be born?What conservation lawis in contradictionwith it ?

23. Quarks inside protons and neutrons are thought to carryfractional charges 2 e, 1 e 3 3 . ..... . .. ....Whyare theynot evidenced inMillikan�s oildrop experiment ?

MODERN PHYSICS www.physicsashok.in 124 Atomic Structure, Radioactivity, Nuclear Fassion and Fusion 1. Distinguish between�excitation� and �ionization� bycollision. 2. According to the theory of electron transitions, the spectral lines froma glowing gas should be �sharp� i.e., of one particularwavelength each. In practice, theyare found to be somewhat �diffuse�, i.e., spread over a small range ofwavelengths. Suggest a reason for this. 3. Can it be concluded from.-decay that electrons exist inside the nucleus ? 4. Why are .-rays emitted only in nuclear processes and not in orbital electron transitions ? 5. What are the principles that are obeyed in filling the orbits of an atom? 6. Howis the radioactivityof an element affectedwhen it forms chemical compounds ? 7. How can Becquerel ray, i.e., the combination of .-, .- and .-rays, be separated ? 8. When a nucleus undergoes .-dcay, is the product atomelectrically neutral ? In .-decay? 9. Do .-decay and .-decay cause a change of element, called transmutation ? 10. Experimental results inradioactivityshowsmallvariations fromthe results predicted bytheory. Explain this. 11. Does the relationE =mc2 suggest that mass can be converted to energy onlywhen it is inmotion ? 12. What is a �thermalneutron� ? 13. Does a nucleus have to be bombardedwith fast or showneutrons in order for it to undergo fission ? 14. Why has it not been possible so far to control the fusion process and obtain usable energy fromit ? 15. An atomhas a continuous distribution ofmass in..... (Thomsonmodel, Rutherfordmodel) but has a highly non-uniformdistributionin ..... (Thomsonmodel,Rutherfordmodel). 16. Which level of the doubly-ionized lithium(Li++)ion has the same energy as the ground state energyof the hydrogen atom? 17. If the a-decayofU238 is allowed fromthe point of viewof energy (the decayproducts have a totalmass less than themass ofU238) what preventsU238 fromdecaying all at once ?Why is its half-life so large ? 18. Can a spectral line belong to both the Lyman andBalmer series ? 19. Although theLymanseries involves transitions to the ground level, andtheBalmer series to the second orbit, the latter was discovered earlier.Why ? 20. Bohr�s principle of quantization of angularmomentumis not a postulate but an essentialcondition. Explain how. 21. Whyare .-particle tracks as observed in a cloud chambermuch shorter than .-particle tracks though they emerge froma radioactive samplewith almost the same speed ? 22. Cathode rays and .-particles are streams of electrons. Inwhat respect do theythen differ fromeach other? 23. When a radioactive substance emits an .-particle its position in the periodic table is lowered bytwo places. Is this true or false ? 24. When Rutherford bombarded a thin foil of gold by .-particles, he found that 1 in 2500 are deflected through verylarge angles.What inference did he drawfromthis result ? 25. Auraniumnucleus (atomic number 92, mass number 238) emits an .-particle and the resultant nucleus

emits a .-particle.What are the atomic andmass numbers of the final nucleus ? 26. Among .-particles, .-particles, protons and neutronswhich have the greatest penetrating power through matter andwhy ? 27. The isotope of hydrogen 31 H (tritium) is radioactive.What would be its decay process and the product? 28. What ismeant by the disposal of radioactivewaste in a nuclear reactor ? 29. .- and .-particles suffer equal and opposite deflections in an external electric field. Is this true or false? 30. It requires infinite time for all the atoms in a radioactive sample to decay,whatever be the half-life of the material. Is this true or false ? 31. What ismean by �enrichment of uranium� ? 32. Which yields greater energy per atom�fission or fussion ? .... per unitmass ? 33. Explain the statement �Themoderator ina nuclear reactor thermalizes the neutrons.�

MODERN PHYSICS www.physicsashok.in 125 34. When is a chain reaction said to be critical ? 35. If a nucleus emits only a .-ray photon, does itsmass number change?Does itsmass change ? 36. Aclassical atombased on ..... (Thomsonmodel, Rutherfordmodel) is doomed to collapse.Why ? 37. What is Bohr�s correspondence principle ? 38. Bohr�s quantization principle, i.e., angularmomentum= nh/2. is a basic lawin nature.Why dowe never speak of quantizationof the angularmomentumof a planet, around the sun ? 39. Aphoton is emitted by a dense star. Scientists say there is a change in the frequency of the photon as it moves awayfromthe star and call the difference infrequcney gravitational shift. Can you explain this? SOLUTION OF THINKING PROBLEMS X-rays 1. X-rays cannot be deflected by electric andmagnetic fields. They are reflected, refracted, diffracted like ordinarylight waves.All these facts lead us to believe that X-rays are electromagneticwaves. 2. The quantumtheory predicts a lower bound (hv =Ve) ofwavelengths ofX-rays produced in anX-rays tube. This is found to be in good agreement withexperimentalobservations. So this supports the quantum theoryof light. 3. No. Infact electrons have associatedwaves.At the accelerating voltage of the experiment thewave behaviour of electrons becomes quite prominent and interference fringes are observed. 4. False, X-rays damage living cells and hence kill the bacteria present inwater. Hence, the water actually becomes safer for drinking. 5. The electronloses energydue to collisionswith the gas atoms. It therefore describes a circlewith decreasing radius, i.e., it spirals inwards. 6. The phenomenon offluorescence consists of absorptionof higher energyphotons and re-radiation of lower energy visible light. This is possiblewith ultraviolet rays as these have greater photon energy than visible light. Infrared rays have lower photon energy, so fluorescence cannot occur. 7. D, B. Alpha particles are heavy and positively charged, and so they are deflected the least to the left according to Fleming�s left-hand rule. The electrons are deflected to the right. 8. Yes. The de Brogliewavelength does not depend on charge, onlyonmomentum. 9. X-rays or .-rays cause ionization bycollision inthe gas atoms. The free electron and ionpairs produced can nowmove and conduct electricity. Thus, the gas becomesmore conducting. 10. Theyhave large atomic numbers and highmelting points. 11. No.The electric fieldwillnot affect the initialvelocity. Itwillonlyproduce an additional acceleratedmotion perpendicular to the initial velociyt.A90º deflection requires that the initial velocity be reduced to zero. 12. (a) No work is done, since the displacement due to the motion is perpendicular to the field. (b)Work iis done, as the displacement howhas a component parallel to the force. 13. (a) The speed does not change since the force is perpendicular to the displacement at everypoint. The field does no work on the particle. The energy and speed of the particle remain unchanged.

(b)When an electric field acts for some time, the particlewill acquire a displacement parallel to the force. Work is done by the field on the particle. The energyand speed of the particlewill change. 14. Due to themagnetic field, the particlewillmove in a circlewith the field as axis.Due to the electric field, a forcewill act along this axis, producing an acceleratedmotion perpendicular to the plane of the circle. The result willbe a helicalpath ofgradually increasing pitch. 15. Their negativelycharged character is shownbythe directionofdeflection inan electric field. Their streaming character is shown by deflection in amagnetic field because a magnetic field can produce deflection only when charged particles are inmotion.

MODERN PHYSICS www.physicsashok.in 126 16. The charged particlewillmove along a helixwinding around the lines of induction. Let us resolve the velocityalong the field and perpendicular to it. The resolved part along the field vII is called orbital velocity. Since the magneticmoment due to the orbitalmotion is opposite to themagnetic field is tends to push the charged particle out of the field, i.e., the charged particle is strongly decelerated and so its drift velocitydecreases, becoming zero inthe case of a sufficientlyhigh field gradient. Fromthis place it begings tomove in the opposite direction. 17. The charged particles followa helicalpath of periodT = 2.m/Bqwhichis independent of the velocityof the particles and pitch p = (2.m/Bq) v cos.. = (2.mv)/Bq when . small. Thus after a path of length p all the particles come down to the same point whatever be the angle of inclination of their initialmotionwith the field. 18. It is absorbed bythe target,which is also the anode of theX-raytube. Subsequently, the electronreturns to the cathode via the external voltage circuit. 19. Less than 1%of the incident electronenergy is actuallyconverted toX-rays. The balance is lost in inelastic collisions be between the electrons and the target atoms. This energy appears as heat in the target. 20. In aCoolide tube, the hardness and intensity of theX-rays can be controlled independently. The hardness is controlled by the applied voltage. The intensity is controlled by the filament temperature, i.e., by the filament current. Such independent control is not possible in the cathode ray tube. 21. Thewavelengthof amatterwave is given by. =h/p.Themomentumofordinarymaterialbodies at ordinary speeds is verylarge and so the associatedwavelength is extremelysmallbecause of the verysmallvalue of h = 6.6 × 10�34 Js. This iswhy thewave nature ofmatter is not apparent in our daily lives. 22. Asingle photoncannot be bornbecause the principle ofconservation ofmomentumwould then be violated. The meson is at rest and so if a single photon is created, that photon must also be at rest to conserve momentum. But it is not possible for a photon to be at rest. This iswhy two photons are bornwhichmove in opposite directions after creation, in conformitywith the principle of conservation ofmomentum. 23. Because they are held together by a strong force, they are not exhibited separately. Atomic Structure, Radioactivity, Nuclear Fission and Fusion 1. When amoving electron collideswith an atom, one orbital electron in the atomabsorbs part or allof the kinetic energyof the incident electron.As a result, the orbital electronmaymove to an outer orbit, of higher energy (excitation), or become completelyfree fromthe attractive field ofthe nucleus (ionization). 2. The gas atoms emitting light due to electron transitions are inmotion. They behave like fixed frequency sources inmotion.Due toDoppler effect, the observed frequency(and hencewavelength) becomes different fromthe emitted frequency.This difference depends onthe velocityofthe atom, caus

ing the small spread of the spectral line. 3. No. The .-particle, although an electron, is actuallycreated at the instant of .-decay and ejected at once. It cannot exist inside the nucleus as its de Broglie wavelength ismuch larger than the dimensions of the nucleus. 4. The energyof a .-rayphotonis of the order ofMeV. Energies of thismagnitude occur innuclear processes but not inorbital electron transitions. 5. The universalprinciple of stability ofa system, that is, �a systemlies instable equilibriumwhen its energy is at the lowest possibelvalue�and Pauli�s exclusion principle, i.e., �no two electrons canhave all their quantum numbers identical�. 6. In no way. Chemical bonds involve onlyorbital electrons, whereas radioactivity is a nuclear process. 7. Bypassing themthrough transverse electric ormagnetic fields. 8. No. In .-decay, the atomic number decreases by2, hence the atomis left with two extra orbital electrons. It therefore has a double negative charge. In .-decay, the atomis left with a net single positive charge.

MODERN PHYSICS www.physicsashok.in 127 9. Yes. In .-decay, the element moves back two places on the periodic table. In .-decay, the element moves forward one place on the periodic table. 10. The lawof radioactive decay is statistical in nature. Hence, individial experimental resultswillshowslight variations.The averages over a large number of experimental results conformexactlywith the theory. 11. No. Here c2 appears only as a constant and does not suggest motion. 12. This is a neutronwithenergyof the order of(3/2) kT,whereTis the absolute temperature of the surroundings and k is Boltzmann�s constant. This follows fromcomparisonwith the law of equipartition of energy as applied to gasmolecules. �Thermalisation� of a neutron brings down its energy froma high value of about (3/2) kT. 13. For fission, theneutronmust be absorbed bythe fissionable nucleus.This is possible onlywith slowneutrons. 14. Fusion occurs onlyat temperatures of the order of 106K. Thismakes it extremely difficult to control fusion processes. 15. Thomsonmodel,Rutherfordmodel. 16. E . Z2 / n2 .When E is constant, Z2 . n2 . Z.n . 3/1 = n/1 . n = 3 17. The emission of a-particles is caused by quantummechanical tunnelling through the repulsive Coulomb barrier. They bounce to and fro in the potentialwell bounded by the barrier before tunnelling through it. Hence the probability of escape is not the same for all the a-particles because all are not born inside the nucleus at the same time. 18. Spectral lines fromhydrogen arise fromthe relation hv = E0(1/n2 � 1/m2) withm> n. For the Lyman series n = 1, m= 2, ...., . . . (hv)max = E0 and (hv)min = (3/4) E0 or 34 E0/h < v < E0/h For the Balmer series, n = 2, m= 3, ....., . . . (hv)max = E0/4 and (hv)min = 5E0 /36 or 5E0/36h < v < E0/4h. Clearly, the same value of v cannot satisfy both the series. Hence, a spectral line cannot belong to both series. 19. Because theBalmer series lies inthe visible region and the Lyman series in the ultraviolet region. 20. Bohr�s principle of quantizationof angularmomentumis seen to be an essential conditionwhile considering de Broglie�smatterwave principle, that is, innature allmoving bodies have an associatedwave.We have p = h/.. In the stationary orbits thewaves associatedwiththe particlemust forma stationarywave. If r is the radius ofthe stationaryorbit, its circumference (equalto 2.r)must be inmultiples ofwavelength..Therefore, 2.r = n. or 2.r = n . h/p or pr = nh/2..Moment ofmomentumis angular momentum. Hence pr is the angularmomentum(L) of the electron. Thus finallywe find that I.= nh/2.. 21. .-particles have greater ionizing power than .-particles and so they lose their energymuch earlier than

.-particles because of collisions. This iswhy their track lengths are shorter than those of .-particles. 22. Theydiffer in respect of their origin. .-particles originate in the nucleuswhen a neutron is converted into a proton, whereas electrons in cathode rays are orbital electrons. 23. Yes. Since .-particles are heliumatoms, their emissionlowers the atomic number by2 andmass number by 4. Elements are arranged in the periodic table according to their atomic number. So emittion of .-particles lowers the position of the element by two places in the periodic table. 24. An atomconsists of a smallcentralizedmass containing positivelycharged particles.Otherwise the atomas awhole in empty. 25. 92 and 234.92U238 � 2He4 � 2 × �1e0 . 92U234

MODERN PHYSICS www.physicsashok.in 128 26. Neutrons, because theyare electricallyneutral and so they do not interactwithmatter electrically. 27. The only possible decay process is .-decay. The decayproduct would be 2He3. 28. In nuclear fission, two nuclideswith Z of the order of 40 to 50 are created. These are highly radioactive, with half-lives or thousands of years. These are called �radioactivewaste�. They have to be disposed of in sealed containerswhich can contain their radioactive emissions. 29. False. .-particles are deflectedmore due to their larger specific charge. 30. True.This follows fromthe exponentialnature ofthe decay. In N = N0e�.t, for N = 0, t = . 31. Naturaluraniumcontains less than1%ofU235mixedwithU238. The latter is not fissionable,while the former is fissionable. The proportion ofU235must be increased artificiallyfor the uraniumto be used in fission. This is called enrichment of uranium. 32. Fission yields greater energyper atom. Fusion yields greater energyper unit mass. 33. The neutrons emitted in fissionmust be slowed down inorder to cause further fission in other nuclei.This is called thermalization, and is performed bythemoderator. 34. When exactly one neutron, of the several produced by the fission of one nucleus, is permitted to cause further fission. This happens in a controlled chain reaction, e.g., ina nuclear reactor. 35. Themass number does not change. Themass is reduced. 36. Thomson�smodel. 37. For large quantumnumbers, quantummechanical results reduce to classical results. 38. Because large n corresponds to a verylarge value at which classical and quantumresults are identical by Bohr�s correspondence principle. 39. As the photonmoves out against strong gravitational attraction its energydecreases and so its frequencyis expected to decrease.This is gravitational shift.

MODERN PHYSICS www.physicsashok.in 129 ASSERTION-REASON (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statment-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True 1. Statement-1 : Electron capture occursmore often than positron emission in heavy elements. Statement-2 : Heavy elements exhibit radioactivity. 2. Statement-1 : In a hydrogen atomenergy of emitted photon corresponding to transition fromn = 2 to n = 1 ismuch greater as compared to transition fromn = . to n = 2. Statement-2 :Wavelengthof photon is directlyproportional to the energyof emitted photon 3. Statement-1 : Ionisation energyof atomic hydrogen is greater than atomic deuterium. Statement-2 : Ionisation energy is directly proportional to reducedmass 4. Statement-1 : For pair production, energy of . ray is greater than 1.02MeV. Statement-2 : In pair production, energy is converted into mass. 5. Statement-1 : The ratio of rate of production (R) of neutrons to the rate of leakage of neutron froma spherical body of 92U235 is directlyproportional to radius (r) Statement-2 : Rate of production of neutron is directly proportional to volume but rate of leakage of neutrons is directlyproportional to area. 6. Statement-1 : The nuclear energycan be obtained bythe nuclear fissionof heavier nuclei aswell as nuclear fusion oflighter nuclei. Statement-2 : The binding energy per nucleonwith increase in atomic number first increases and then decreases. 7. Statement-1 : Asmallmetal ball is suspended in a uniformelectric fieldwith an insulated thread. If high energyX-ray beamfalls on the ball, the ballwill be deflected in the direction of electric field. Statement-2 :Wavelength of L. X-raymust be greater then the wavelength ofK. X-ray for the same material. 8. Statement-1 : The difference in the frequencies ofseries limit ofLymanseries andBalmer series is equal to the frequency of the first line of the Lyman series. Statement-2 :Difference inenergyof two atomic levels is proportionalto the energyofemitted or absorbed photon. 9. Statement-1 :Work function of aluminiumis 4.2 eV. If two photons of eachof energy 2.5 eVstrike on an electron ofaluminium, the electron is not emitted. Statement-2 : In photoelectric effect, electron is emitted onlyif energyofeach of incident photonis greater than thework function. 10. Statement-1 : Ifthe acceleratingpotentialinanX-raytube is increased, thewavelengthsofthe characteristic X-rays do not change. Statement-2 :When an electron beamstrikes the target in anX-ray tube, part of the kinetic energy is converted intoX-rayenergy. [JEE, 07]

MODERN PHYSICS www.physicsashok.in 130 Match the Column 1. Column I Column II (A) Work function of copper is 4 eV. If two (P) �13.6 Z2/n2eV photons each of energy 2.5 eV strike an electron of copper emission of electrons (B) Cathode rays get deflected by (Q) 13.6 Z2 2 2 1 2 1 1 n n . . . . . . . eV (C) Ionisation energy of H like atom is (R) both electric and magnetic field (S) 1H1 (D) Greater wavelength in transition from n = 2 (T) 1H3 to n = 1 is for (U) Not possible (V) Possible 2. Column I Column II (A) Radius of orbit depend on principal quantum (P) Increase number as (B) Due to orbital motion of electron, (Q) decrease Magnetic field arises at the centre of Nucleus is proportional to principal quantum no. as (R) is proportional to 1/n2 (C) If electron is going from lower energy level to higher energy level then velocity of electron will (S) is proportional to n2 (D) If electron is going from lower energy level to higher energy level, then total energy of (T) is proportional to 1/n5 electron will 3. Column I Column II (A) Rate of disintegration, i.e. dN/dt is (P) Greater than Half life proportional to (B) Mean life of radioactive substance is (Q) Less than Half life (C) Intensity I of . ray of initial intensity I0 after (R) Number of atoms of parent radioactive transversing the thickness of x of the absorber is substance still undecayed at time t (related as) (D) The radioactive decay rate is not affected by (S) I = I0/x (T) I = I0e�.x (U) Temperature, pressure, volume 4. Column I Column II (A) Binding energy per nucleon for middle order (P) Optical Model or elements is (B) Nuclear force depends on (Q) Shell model (C) For nuclear fission Z2 A is (R) 8.8 MeV (D) Magic numbers are 2, 8, 20, 28, 50, 82, 126 (S) 2.5 eV explained by (T) Charges of Nucleons (U) Spin of Nucleons (V) Greater than 15 (W) Less than 15 5. Column I Column II (A) Radius of orbit is related with atomic number (Z) (P) is proportional to Z

(B) Current associated due to orbital motion (Q) is inversely proportion to Z electron with atomic number (Z) (C) Magnetic field at the centre due to orbital (R) is proportional to Z2 motion of electron related with Z. (D) Velocity of an electron related with atomic (S) is proportional to Z3 number (Z)

MODERN PHYSICS www.physicsashok.in 131 6. Match the followingColumns [JEE, 06] Column I Column II (A) Nuclear fusion (P) Converts somematter into energy (B) Nuclear fission (Q) Generallyoccurs for nucleiwith lowatomic number (C) .-decay (R) Generallyoccurs for nucleiwith higher atomic number (D) Exothermic nuclear reaction (S) Essentially proceeds byweek nuclear forces 7. In the following, column I lists some physical quantities&the column II gives approx, energy values associated with some of them. Choose the appropriate value of energy fromcolumn II for each of the physical quantities in column I andwrite the corresponding letterA, B, Cetc. against the number (i), (ii), (iii), etc. of the physical quantityin the answer book. In your answer, the sequence of column I should be maintained. Column I Column II [JEE, 97] (A) Energyofthermalneutrons (P) 0.025 eV (B) EnergyofX-rays (Q) 0.5 eV (C) Binding energyper nucleon (R) 3 eV (D) Photoelectric threshold ofmetal (S) 20 eV (T) 10 keV (U) 8MeV 8. Some laws/processes are given inColumnI.Match thesewiththe physicalphenomena giveninColumn II. Column I Column II (A) Transition between two atomic energylevels (P) CharacteristicX-rays [JEE, 07] (B) electron emissionfromamaterial (Q) Photoelectric effect (C) Mosley�s law (R) Hydrogen spectrum (D) Change of photonenergyinto kinetic (S) .-decay energyof electrons. 9. Column- II gives certainsystemsundergoing aprocess.Column-I suggests changes insomeofthe parameters related to the system.Match the statements inColumn I to the appropriate process(es) fromColumn-II. Column I Column II [JEE, 09] (A) The energy ofthe systemis increased (P) System: Acapacitor, initiallyuncharged (B) Mechanical energyis provided to the system, Process : It is connected to a battery which is converted into energy of random (Q) System: Agas inan adiabatic container fitted motion ofits parts with an adiabatic piston (C) Internal energyof the systemis converted into Process :The gas is compressed by pushing itsmechanical energy. the piston (D) Mass of the systemis decreased (R) System: Agas in a girid container Process :The gas gets cooled due to colder atmosphere surrounding it (S) System: Aheavynucleus, initiallyat rest Process :The nucleus fissions into two fragments of nearly equalmasses and some neutrons are emitted. (T) System:Aresistivewire loop Process :The loop is placed in a time varrying magnetic fieldperpendicular to its plane

MODERN PHYSICS www.physicsashok.in 132 Level � 1 1. An energy of 24.6 eV is required to remove one of the electrons from the neutral helium atom. The energy (in eV) required to remove both the electron from a neutral helium atom is (a) 38.2 (b) 49.2 (c) 51.8 (d) 79.0 2. The K X . ray . emission line of tungsten occurs at . = 0.021 nm. The energy difference between K and L levels in this atom is about (a) 0.51 MeV (b) 1.2 MeV (c) 59 keV (c) 136 eV 3. The electron in a hydrogen atom makes a transition n1 .n2 where n1 and n2 are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is 8 times that in the final state. The possible values of n1 and n2 are (a) n1 = 4, n2 = 1 (b) n1 = 8, n2 = 2 (c) n1 = 8, n2 = 1 (d) n1 = 6, n2 = 3 4. X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from (a) 0 to . (b) min . to . where min . > 0 (c) 0 to max . where max . < . (d) min . to max . where 0 < min . < max . < . 5. A particle of mass M at rest decays into two particles of masses m1 and m2, having non zero velocities. The ratio of the de Broglie wavelengths of the particles, 1 2 . . , is (a) 21 mm (b) 12 mm (c) 1.0 (d) 12 mm 6. The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements is true ? (a) Its kinetic energy increases and its potential and total energies decrease. (b) Its kinetic energy decreases, potential energy increases and its total energy remains the same. (c) Its kinetic and total energies decrease and its potential energy increases. (d) Its kinetic, potential and total energies decrease. 7. Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength . (given in terms of the Rydberg constant R for the hydrogen atom) equal to (a) 9/5R (b) 36/5R (c) 18/5R (d) 4/R 8. Electrons with energy 80 keV are incident on the tungsten target of an X-ray

tube. K shell electrons of tungsten have � 72.5 keV energy. X-rays emitted by the tube contain (a) a continuous X-ray spectrum (Bresmsstrahlung) with a minimum wavelength of about 0.155A. (b) a continuous X-ray spectrum (Bremsstrahlung) with all wave-lengths (c) the characteristic X-ray spectrum of tungsten (d) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of about 0.155A.and the characteristics X-rays spectrum of tungsten. 9. For a photoelectric cell, the graph in Figure. showing the variation of the cut-off voltage V0 with frequency (v) of incident light is O v V0 (a) O v V0 (b) O v V0 (c) O v V0 (d)

MODERN PHYSICS www.physicsashok.in 133 10. Monochromatic light of frequency v1 irradiates a photocell and the stopping potential is found to be V1. What is the new stopping potential of the cell if it is irradiated by monochromatic light of frequency v2? (a) 1 .v2 v1. eV . h . (b) . . 1 v2 v1 eV . h . (c) . . 1 v1 v2 eV . h . (d) . . 1 v1 v2 eV . h . 11. When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for photoelectric current is V0/2. When the same surface is illuminated by monochromatic high of frequency v/2, the stopping potential is V0. The threshold frequency for photoelectric emission is (a) 32v (b) 23v (c) 53v (d) 35v 12. The energy of a photon of frequency v is E = hv and the momentum of a photon of wavelength . is p = h / . . From this statement one may conclude that the wave velocity of light is equal to (a) 3 x 108 ms�1 (b) .. (c) .p (d) 2 p. .. . . .. . . 13. When a centimeter thick surface is illuminated with light of wavelength . , the stopping potential is V. When the same surface is illuminated by light of wavelength 2 . , the stopping potential is V/3. The threshold wavelength for the surface is (a) 3 4 . (b) 4 . (c) 6 . (d) 3 8 . 14. A star of mass M0, radius R0 contracts to radius R. Energy radiated by the star assuming uniform density in each case while temperature remains unchanged is (a) .0 c. (b) . . .. . . .

.

. ..

.

. ..

.

. . . 2 0 0 Rc 1 R (c) . .. . . .. . . . . 0 0 Rc 1 R (d) . . .. . . .. . .. . . .. . . . . 3 0 0 Rc 1 R 15. A sensor is exposed for time t to a lamp of power P placed at a distance . . The sensor has an opening that is 4d in diameter. Assuming all energy of the lamp is given off as light, the number of photons entering the sensor if the wavelength of light is . is (l >> d) (a) 2 2 hc N P d t . . . (b) 22 hc N 4P d t . . . (c) 2 2 4hc N P d t . . . (d) 2 2 16hc N P d t . .

. 16. If elements of quantum number greater than n were not allowed, the number of possible elements in nature would have been (a) n .n 1. 21 . (b) . . 2 2 n n 1 . . . . . .. (c) n.n 1. .2n 1. 61 . . (d) n.n 1. .2n 1. 31 . . 17. An electron is lying initially in the n = 4 excited state. the electron de-excites itself to go to n = 1 state directly emitting a photon of frequency v41 . If the same electron first de-excites to n = 3 state by emitting a photon of frequency v43 and then goes from n = 3 to n = 1 state by emitting a photon of frequency v31 , then (a) 41 43 31 . . . . . (b) 41 43 31 . . . . . (c) .41 . .43 . 2.31 (d) Data Insufficient 18. A photon of energy 10.2 eV corresponds to light of wavelength 0 . . Due to an electron transition from n = 2 to n = 1 in a hydrogen atom, light of wavelength . is emitted. If we take into account the recoil of the atom when the photon is emitted, (a) . = 0 . (b) 0 . . . (c) 0 . . . (d) the data is not sufficient to reach a conclusion.

MODERN PHYSICS www.physicsashok.in 134 19. When an electron moving at a high speed strikes a metal surface, which of the following are possible? (a) The entire energy of the electron may be converted into an X-ray photon. (b) Any fraction of the energy of the electron may be converted into an X-ray photon. (c) The entire energy of the electron may get converted to heat. (d) The electron may undergo elastic collision with the metal surface. 20. A star converts all of its 2He4 nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is (given mass of nucleus = 4.0026 amu, mass of oxygen nucleus = 15.994 amu) (a) 7.26 MeV (b) 7 MeV (c) 15.252 MeV (d) 23.9 MeV 21. The graph showing the energy spectrum of . particles is : (a) E n(E)Y X (b) E n(E)Y X (c) E n(E)Y X (d) E n(E) 22. Binding energy per nucleon of 1H2 and 2He4 are 1.1 eV and 7.0 MeV respectively. Energy released in the process 1H2 + 1H2 = 2He4 is : (a) 20.8 MeV (b) 16.6 MeV (c) 25.2 MeV (d) 23.6 MeV 23. Two electrons are moving with the same speed V. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field . After sometime if the de Broglie wavelength of the two are .1 and .2 then : (a) .1 = .2 (b) .1 > .2 (c) .1 > .2 (d) .1 > .2 or .1 < .2 24. In a characteristic X-ray spectra of some atom superimposed on a continuous X-ray spectra : P Q .min . Relation intensity (a) P represents K. line (b) Q represents K. line (c) Q and P represents K. and K. lines respectively (d) Position of K. and K. depend on the particular atom 25. Difference between nth and (n + 1)th Bohr�s radius of �H� atom is equal to its (n �1)th Bohr�s radius. The value of n is (a) 1 (b) 2 (c) 3 (d) 4 26. A hydrogen atom is in an excited state of principal quantum number n. It emits a photon of wavelength . while returning to the ground state. The value of n is :

(a) .R(.R .1) (b) ( R 1) R . . . (c) R R 1 . . . (d) (R 1) . . 27. The binding energies of nuclei X and Y are E1 and E2 respectively. Two atoms of X fuse to give one atom of Y and an energy Q is released. Then : (a) Q = 2E1 � E2 (b) Q = E2 � E1 (c) Q < 2E1 � E2 (d) Q > E2 � 2E1 28. Two radioactive materials x1 and x2 have decay constant 10 . and . respectively. Initially they have the same number of nuclei. The ratio of the number of nuclei x1 to that of x2 will be 1/e after time : (a) 1/10. (b) 1/11. (c) 11/10. (d) 1/9.

MODERN PHYSICS www.physicsashok.in 135 29. At t = 0 activity of radioactive substance is 1600 Bq and t = 8 sec activity remains 100 Bq. The activity at t = 2 sec is : (a) 200 Bq (b) 400 Bq (c) 600 Bq (d) 800 Bq 30. There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half life of neutrons is 700 s. The fraction of neutrons will decay before they travel a distance 10 m. mn = 1.675 × 10�27 kg : (a) 3.96 × 10�5 (b) 3.96 × 10�6 (c) 2.96 × 10�4 (d) None 31. The count rate for 10 gram radioactive material was measured time (hrs) at different times and this had been shown in figure scale given. The half life of material and the total count in the first half value period respectively are : (a) 4 hrs. and 9000 approximately (b) 3 hrs and 14100 approximately (c) 3 hrs and 235 approximately 75 50 25 3 6 9 100 12.5 (d) 10 hrs and 157 approximately time in hr. 32. Assuming that about 200 MeV energy is released per fission of 92U235 nuclei. What would be mass of 92U235 consumed per day in the fission of reactor of power 1 MW approximately ? (a) 10 kg (b) 100 kg (c) 1 gram (d) 10�2 gm 33. The energy, the magnitude of linear momentum and orbital radius of an electron in a hydrogen atom corresponding to the quantum number n are E, P and r, according to the Bohr�s theory of hydrogen atom: (a) EPr is proportional to 1/n (b) P/E is proportional to n0 (c) Er is not constant for all orbits (d) Pr is proportional to n. 34. An electron is excited from a lower energy state to a higher energy state in a hydrogen atom. Which of the following quantities decrease in the excitation. (a) Potential enrgy (b) Angular speed (c) Kinetic energy (d) Angular momentum 35. The correct statement is l are : (a) density of nucleus is independent of mass number (A) (b) Radius of nucleus increases with mass number (A) (c) Mass of nucleus is directly proportional to mass number (A). (d) Density of nucleus is directly proportional to mass number. 36. A hydrogen like atom of atomic number Z is an excited state of quantum number 2n. It can emit a maximum energy photon of 204. eV. It makes a transition to quantum state n, a photon of energy 40.8 eV is emitted, then (a) Z = 2 (b) Z = 4 (c) n = 1 (d) n = 2 37. An electron with kinetic energy varying from 5 eV to 50 eV is incident on a hydrogen atom in its ground state. The collision : (a) may be elastic (b) may be partially elastic (c) must be completely inelastic (d) from zero to 13.6 eV be elastic and more than 27.2 eV be inelastic. 38. The wavelength of first Balmer line for 1H1 , 1H2 and 2He4 and .1, .2 and .3

respectively. The correct statements are (a) .2 ...1 (b) .3 ...2 (c) .1 ...2 (d) .1 ...2 > .3 39. Regarding X ray spectrum which of the following statement (a) The characteristic X ray spectrum is emitted due to excitation of inner electrons of atom (b) Wavelength of characteristic spectrum depend on the potential difference across the tube. (c) Wavelength of continuous spectrum is dependent on the potential difference across tube (d) None of these

MODERN PHYSICS www.physicsashok.in 136 40. If the wavelength of light in an experiment on photo electric effect is doubled : (a) The photoelectric emission will not take place. (b) The photoemission may or may not take place. (c) The stopping potential will increase (d) The stopping potential will decrease under the condition that energy of photon of doubled.Wavelength is more than work function of metal. 41. the binding energy per nucleon is : (a) Maximum for middle order element (b) Minimum for lighter elements (c) Binding energy per nucleon suddenly increases for some mass number called magic numbers. (d) Binding energy per nucleon is minimum for middle order elements 42. When Z is doubled in an atom, which of the following statement are consistent with Bohr�s theory : (a) Energy of a state is double. (b) Radius of an orbit is double. (c) Velocity of electrons in an orbit is doubled (d) Radius of orbit is halved. 43. Photons of wavelength 6620 Å are incident normally on a perfectly reflecting screen. Calculate the number of photons per second falling on the screen as total power of photons such that the exerted force is 1N : (a) 5 × 1026 (b) 5 × 1025 (c) 1.5 × 108 (d) None of these 44. The energy of . particles emitted by 210Po is 5.3 MeV. What mass 210Po is needed to power a thermoelectric cell of 13 watt output, What would be power output after 1 year : (The half life of 210Po is 138 days) (a) 8.85 × 10�2 gram (b) 0.159 watt (c) 0.179 watt (d) 8.85 × 10�4 gram 45. The atomic masses of 7N15, 8O15 and 8O16 are respectively 15.0001 a.m.u., 15.0030 a.m.u. and 15.9949 a.m.u. Then : (a) Binding energy per nucleon in 8O16 is 7.97 MeV (b) Energy is needed to remove one proton 8O16 is 12.13 MeV (c) Energy needed to remove one proton from 8O16 is 15.61 MeV. (d) All the above 46. A sample contains 10�2 kg each of two substances A and B with half lives 4 sec and 8 sec respectively. Their weights are in the ratio of 1 : 2. The amounts of a and B after an interval of 16 sec. (a) 6.25 × 10�4 kg (b) 12.5 × 10�4 kg (c) 2.5 × 10�3 kg (d) 1.25 × 10�5 kg 47. The wavelength and frequency of photons in transition 1,2 and 3 for H like atom are .1, .2, .3 and .1, .2, .3. then : 123 CBA .1 .2 .3 (a) .3 = .1 + .2 (b) .3 = .1 + .2 (c) 1 2 3 1 2 . . . .

. . . (d) 1 2 3 1 2 . . . . . . . 48. Which of the following pair constitute very similar radiations ? (a) Hard U.V. ray and soft X ray. (b) Soft U.V. ray and hard X ray (c) Very hard X ray and low frequency Y ray (d) Soft X ray and Y ray 49. The correct option are : (a) In uranium ore, the ratio of U235 to U238 is 1 : 40 (b) Critical mass of uranium is 10 kg (c) 92U235 : 92U238 = 1 : 4 (d) All the above

MODERN PHYSICS www.physicsashok.in 137 50. A star initially has 1040 deutrons. It produces energy via, the processes 1H2 + 1H2 ... 1H3 + p & 1H2 + 1H3 ... 2He4 + n. If the average power radiated bythe star is 1016W, the deuteron supplyof the star is exhausted in a time of the order of : [JEE, 93] (A) 106 sec (B) 108 sec (C) 1012 sec (D) 1016 sec 51(i). Fast neutrons can easily be slowed down by : (A) the use of lead shielding (B) passing themthroughwater (C) elastic collisionswith heavynuclei (D) applying a strong electric field (ii). Consider .�particles, .�particles & .-rays, each having an energy of 0.5 MeV. Increasing order of penetrating powers, the radiations are : [JEE, 94] (A) ., ., . (B) ., ., . (C) ., ., . (D) ., ., . 52. Which ofthe following statement(s) is (are) correct ? [JEE, 94] (A) The rest mass of a stable nucleus is less than the sumof the rest masses of its separated nucleons. (B) The rest mass of a stable nucleus is greater than the sumof the rest masses of its separated nucleons. (C) Innuclear fusion, energy is released byfusion two nuclei ofmediummass (approximately100 amu). (D) In nuclear fission, energyis released by fragmentation of a very heavy nucleus. 53. The binding energy per nucleon of 16O is 7.97 MeV & that of 17O is 7.75 MeV. The energy in MeV required to remove a neutron from17Ois : [JEE, 95] (A) 3.52 (B) 3.64 (C) 4.23 (D) 7.86 54. Themaximumkinetic energyofphotoelectrons emitted froma surfacewhenphotons of energy6 eVfallon it is 4 eV. the stopping potential inVolts is : [JEE, 97] (A) 2 (B) 4 (C) 6 (D) 10 55. Select the correct alternative(s). [JEE, 98] (i) Let mp be themass of a proton,mn themass of a neutron,M1 themass of a 20 10 Ne nucleus&M2 themass of a 40 20Ca nucleus. Then : (A) M2 = 2M1 (B) M2 > 2M1 (C) M2 < 2M1 (D) M1 < 10(mn + mp) (ii) The half-life of 131I is 8 days. Given a sample of 131I at time t = 0, we can assert that : (A) no nucleuswill decay before t = 4 days (B) no nucleuswill decay before t = 8 days (C) all nucleiwill decay before t = 16 days (D) a given nucleusmaydecayat anytime after t = 0 56(a). Binding energyper nucleon vs.mass number curve for nuclei is shownin the figure.W,X,Yand Zare four nuclei indicated on the curve. The process that would release energyis [JEE, 99] 30 60 90 120 Z Y X W 8.5 8.0 7.5 5.0 Mass Number of Nuclei in MeV

Binding Energy/nucleon (A)Y. 2Z (B)W. X + Z (C) W. 2Y (D) X .Y+ Z

MODERN PHYSICS www.physicsashok.in 138 (b) Order ofmagnitude of density ofUraniumnucleus is, [mp = 1.67 × 10�27 kg] (A) 1020 kg/m3 (B) 1017 kg/m3 (C) 1014 kg/m3 (D) 1011 kg/m3 (c) 22Ne nucleus, after absorbing energy, decays into two .-particles and an unknown nucleus.The unknown nucleus is (A) nitrogen (B) carbon (C) boron (D) oxygen (d) Which of the following is a correct statement ? (A) Beta rays are same as cathode rays (B) Gamma rays are high energyneutrons. (C)Alpha particles are singlyionized heliumatoms (D) Protons and neutrons have exactlythe samemass (E)None (e) The half-life period of a radioactive elementXis same as themean-life time ofanother radioactive element Y. Initiallyboth of themhave the same number ofatoms.Then (A) X&Yhave the same decay rate initially (B) X &Ydecay at the same rate always. (C)Ywilldecay at a faster rate thanX (D) X willdecay at a faster rate thanY 57. Aparticle ofmassMat rest decays into two particles ofmassesm1 andm2, having non-zero velocities. The radio of the de-Brogliewavelengths of the particles, .1/.2, is [JEE, 99] (A)m1/m2 (B)m2/m1 (C) 1.0 (D) 2 1 m / m 58.(a) Imagine an atommade up of a proton and a hypotherical particle of double themass of the electron but having the same charge as the electron.Apply theBohr atommodel and consider allpossible transitions of this hypotheticalparticle to the first excited level. The longest wavelength photon that willbe emitted has wavelength ..(given in terms ofthe Rydberg constant R for the hydrogen atom) equal to [JEE, 2000 Scr.] (A) 9/(5R) (B) 36/(5R) (C) 18/(5R) (D) 4/R (b). The electronin a hydrogen atommakes a transition froman excited state to the ground state.Whichof the following statements is true ? [JEE, 2000 Scr.] (A) Its kinetic energy increases and its potential and total energies decrease. (B) Its kinetic energydecreases, potential energyincreases and its total energyremains the same. (C) Its kinetic and total energies decrease and its potential energy increases. (D) Its kinetic, potential and total energies decrease. 59. The potentialdifference applied to anX-ray tube is 5 kV and the current through it is 3.2mA. Then the number of electrons striking the target per second is [JEE, 02 Scr.] (A) 2 × 1016 (B) 5 × 1016 (C) 1 × 1017 (D) 4 × 1015 60. A Hydrogen atom and Li++ ion are both in the second excited state. If lH and lLi are their respective electronic angularmomenta, and EH and ELi their respective energies, then [JEE, 02 Scr.] (A) lH > lLi and |EH| > |ELi| (B) lH = lLi and |EH| < |ELi| (C) lH = lLi and |EH| > |ELi| (D) lH < lLi and |EH| < |ELi| 61. Givena sample ofRadium-226 having half-life of 4 days. Find the probability, a nucleus disintegrateswithin 2 half lives. [JEE, 06] (A) 1 (B) 1/2 (C) 3/4 (D) 1/4

MODERN PHYSICS www.physicsashok.in 139 62. The graph between 1/. and stopping potential (V) of three metals havingwork functions .1, .2 and .3 is anexperiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct? [Here . is thewavelength of the incident ray]. V 0.001 0.002 0.004 1/ nm�1 metal 1 metal 2 metal 3 (A) Ratio ofwork functions .1 : .2 : .3 = 1 : 2 : 4 (B) Ratio ofwork functions .1 : .2 : .3 = 4 : 2 : 1 (C) tan ..is directly proportional to hc/e, where h is Planck�s constant and c is the speed of light (D) The violet colour light can eject photoelectrons frommetals 2 and 3. [JEE, 06] 63. In the options given below, let E denote the rest mass energy of a nucleus and n a neutron. The correct option is : [JEE, 07] (A) .236 . .137 . .97 . 92 53 39 E U . E I . E Y . 2E(n) (B) .236 . .137 . .97 . 92 53 39 E U . E I . E Y . 2E(n) (C) . 236 . .140 . .94 . 92 56 36 E U . E Ba . E Kr . 2E(n) (D) . 236 . .140 . .94 . 92 56 36 E U . E Ba . E Kr . 2E(n) 64. The largestwavelengthinthe ultraviolet regionofthe hydrogenspectrumis 122 nm.The smallestwavelength in the infrared region of the hydrogen spectrum(to the nearest integer) is [JEE, 07] (A) 802 nm (B) 823 nm (C) 1882 nm (D) 1648 nm 65. Electronswith de-Brogliewavelength . fallon the target in anX-ray tube. The cut-offwavelength of the emittedX-rays is [JEE, 07] (A) 2 0 2mc h. . . (B) 0 2h mc . . (C) 2 2 2 0 2 2m c h . . . (D) .0 = . 66. Which one of the following statements isWRONGin the context ofX-rays generated fromaX-raytube? (A)Wavelength of characteristicX-rays decreaseswhen the atomic number ofthe target increases (B) Cut-offwavelengthof the continuousX-rays depends on the atomic number of the target. [JEE, 08] (C) Intensityof the characteristicX-rays depends on the electricalpower given to theX-rays tube. (D) Cut-offwavelengthof the continuousX-rays depends on the energyof the electrons intheX-rays tube.

67. Assume that the nuclear binding energyper nucleon(B/A) versusmass number (A) is as shown inthe figure.Use this plot to choose the correct choice(s) given below: Figure 100 200 A 02468 B/A (A) Fusionof two nucleiwithmass numbers lying in the range of 1 <A< 50 will release energy. (B) Fusionof two nucleiwithmass numbers lying inthe range of 51 <A< 100 will release energy. (C) Fission of a nucleus lying inthemass range of100 <A< 200will release energywhen broken into two equalfragments. (D) Fission ofa nucleus lying inthemass range of 200 <A< 260will release energywhenbroken into two equalfragments. [JEE, 08] 68. The quantumnumber n of the state finally populated inHe+ ions is [JEE, 08] (A) 2 (B) 3 (C) 4 (D) 5

MODERN PHYSICS www.physicsashok.in 140 69. Thewavelength oflight emitted in the visible region byHe+ ions after collisionswithHatoms is [JEE, 08] (A) 6.5 × 10�7 m (B) 5.6 × 10�7 m (C) 4.8 × 10�7 m (D) 4.0 × 10�7 m 70. The ratio of the kinetic energy of the n = 2 electron for theHatomto that ofHe+ ion is [JEE, 08] (A) 14 (B) 12 (C) 1 (D) 2 71. Aradioactive sampel S1 having an activity of 5µCi has twice the number of nuclei as another sample S2 which has an activity of 10 µCi. The half lives of S1 and S2 can be : [JEE, 08] (A) 20 years and 5 years, respectively (B) 20 years and 10 years, respectively (C) 10 years each (D) 5 years each 72. Photoelectric effect experiments are performed using three different metal plates p, q and r havingwork functions .p = 2.0 eV, .q = 2.5 eVand .r = 3.0 eV, respectively.Alight beamcontaining wavelengths of 550 nm, 450 nmand 350 nmwith equal intensities illuminates each of the plates. The correct I�Vgraph for the experiment is [JEE, 09] (A) I pqrV (B) I p q r V (C) I pqr V (D) I r q p V Passage PASSAGE : 1 Figure 1 below depicts three hypothetical atoms. Energy levels are represented as horizontal segments. The distance between the segments is representative of the energy difference between the various levels . All possible transitions between energy levels are indicated by arrows Atom #1 Atom # 2 Atom # 3 Scientists can observe the spectral lines of atoms that are dominant in far-away galaxies. Due to the speed at which these galaxies are travelling, these lines are shifted, but their pattern remains the same. This allows researchers to use the spectral pattern to determine which atoms they are seeing

. Table 1 below shows spectroscopic measurements made by researchers trying to determine the atomic makeup of a particular faraway galaxy. Light energy is not measured directly, but rather is determined from measuring the frequency of light. Which is proportional to the energy.

MODERN PHYSICS www.physicsashok.in 141 Table 1 Frequencies Measured 868440 880570 879910 856390 1. For each of three hypothetical atoms (Atom 1, Atom 2 and Atom 3), Figure 1 depicts the (A) number of electrons and the amount of energy the atom contains (B) distance an electron travels from one part of the atom to another (C) energy released by the atom as an electron as it moves from one energy state to another (D) frequency with which the atom�s electrons move from one energy state to another 2. In which of the three hypothetical atoms depicted in Figure 1 does the energy of the light released by the atom vary the least (A) Atom 1 (B) Atom 2 (C) Atom 3 (D) It is impossible to tell 3. Scientist observing an actual atom similar to hypothetical Atom 1 in the figure might observe � (A) three spectral lines close together and two other spectral lines close together (B) light blinking at six different frequencies (C) a much brighter light emanating from one electron than from any other. (D) four distinct spectral lines emanating from six different electrons. 4. Based on the spectroscopic measurements shown in Table 1, which of the atoms in Figure 1 (Atom 1, Atom2, or Atom 3) is most similar to the one the scientists were observing, and why ? (A) Atom 2, because it contains four different energy levels (B) Atom 3, because it contains four different energy levels (C) Atom 1, because the frequencies listed in Table 1 indicate a high level of atomic activity. (D) Atom 3, because there is a comparatively small difference between exactly two of the four frequencies listed in Table 1 5. The laws of atomic physics prohibit electron movements between certain energy states. In atomic physics. these prohibitions are called �forbidden transitions.� Based on Figure 1, which of the following is most accurate ? (A) Atom 2 has the same number of forbidden transitions than Atom 3 (B) Atom 2 has more forbidden transitions than atom 3 (C) Atom 3 has the same number of forbidden transitions as Atom 1 (D) Atom 1 has fewer forbidden transitions than Atom 2 PASSAGE : 2 In quantum mechanics, some quantities are discrete and cannot be continuous. One of these quantities is the energy. Energy can only take certain values � E1, E2, E3, E4......., which are called energy levels. The energy cannot take any values between E1 and E2, or E2 and E3 or E3 and E4 etc. Certain transitions from one energy level to another result in the emission of a photon of radiation, whereas others can only take place if a photon is absorbed. The energy levels in a newly discovered gas are expressed as:

2 1 n 2 E E z n . . in which �E1z2 is the ground state energy. Take z = 1 for simplicity, but do not assume that the gas is hydrogen. An experiment is designed to measure the energy as a functions of the level. The results obtained are as follows : n En(eV) 2 �144 3 �64 4 �36 6. The ionization energy of the gas must be (A) 244 eV (B) 576 eV (C) 144 eV (D) +13.6 eV 7. The ground state energy is � (A) �144 eV (B) + 144 eV (C) �244 eV (D) none of the above

PASSAGE : 3 The power per unit area reaching the Earth�s surface from the sun, averaged over 24 hours, is 0.2 kW/m2. This solar energy can be converted directly into electrical energy via the photoelectric effect. For example, in the photoelectrical cathode (emitter) incidentlight Anode (collector) photocell A Ammeter cell shown in figure 1, a cathode emits electrons when PASSAGE : 3 The power per unit area reaching the Earth�s surface from the sun, averaged over 24 hours, is 0.2 kW/m2. This solar energy can be converted directly into electrical energy via the photoelectric effect. For example, in the photoelectrical cathode (emitter) incidentlight Anode (collector) photocell A Ammeter cell shown in figure 1, a cathode emits electrons when kinetic energy of Emax will not be able to reach the anode therefore the current will stop altogether. The value of this stopping voltage is dependent only on Emax . 11. The most efficient modern photovoltaic cells can covert the Sun�s energy into electrical energy with an efficiency of 35%. Approximately what area would have to be covered by such cells in order to supply a household with 20 kW-hourse of electrical energy per day � (A) 0.5 m2 (B) 12 m2 (C) 285 m2 (D) 6850 m2 12. Light intensity is defined as the energy flowing per unit area per unit time for an area perpendicular to the direction of energy flow. In an experiment, the frequency of light incident on the cathode of a MODERN PHYSICS 8. Which of the following shapes is most likely to represent the graph of En versus 1/n2 ? (A) (B) (C) (D) 9. A transition from the n = 2 state to the n = 3 state results : (A) in emission of a photon of energy 144 eV (B) in emission of a photon of energy 80 eV (C) in emission of an ultraviolet photon 10. A transition from the n = 4 state to the n = 3 state results : (A) in emission of a photon of energy 28 eV (C) in emission of an infrared photon (D) only accomplished if a photon is absorbed (B) in emission of a photon of energy 13.6 eV (D) only accomplished if a photon is absorbed. electrons travel to the anode, and a small electric current flows. An electron w

ithin the cathode requires a minimum energy to break free from the cathode surface. This minimum energy is known as the work function, W, and is a constant intrinsic to the material of which the cathode is composed. An individual photonincident on the cathode collides with an electron and is absorbed, transferring all of its energy to the electron. The energy of each incident photon is given by Ep = hf, where f is the frequency of incident light and h is Planck�s constant. If Ep is less than W, then no electrons will be ejected from the cathode at all. The maximum kinetic energy Emax� of an electron liberated from the cathode is given by illuminated by light of a high enough frequency. The ejected

:E =E �W

max p

A voltage source can be connected across the photoelectric cell to oppose the current flow. At a critical applied voltage, called the stopping voltage, even an electron ejected from the cathode with a

photoelectric cell is held constant, but the intensity is varied. As the intensity of the incident light is increased, the stopping voltage. it increases.

(A) increases, because more electrons are ejected from the cathode as the number of photons striking (B) remains the same, because the energy supplied to one electron depends only on the energy of an individual photon. (C) increases, because the electrons in the cathode absorb more energy per unit time (D) remains the same, because each incident photon shares its energy between several electrons in the cathode. 13. The behavior of light is sometimes explained in terms of particles and sometimes in terms of waves. Which of the following CANNOT be explained by a theory that refers to light in terms of waves alone � (A) Current flow in a photoelectric cell can be stopped by reducing the intensity of the incident light while maintaining the same frequency. (B) An electron requires energy to escape from the surface of a photosensitive cathode. (C) Current flow in a photoelectric cell can be stopped by reducing the frequency of the incident light while maintaining the same intensity. (D) The angle through which light is refracted when it moves from one medium to another is a function of frequency, rather than intensity. www.physicsashok.in 142

MODERN PHYSICS www.physicsashok.in 143 14. Under which of the following conditions will the stopping voltage across a photoelectric cell be greatest- (A) The wavelength of the incident light is short, and the work function of the cathode material is low. (B) The wavelength of the incident light is short, and the work function of the cathode material is high. (C) The wavelength of incident light is long, and the work function of the cathode is low. (D) The wavelength of the incident light is long, and the work function of the cathode material is high. PASSAGE : 4 An x- ray tube consists of two metal electrodes, a heated filament cathode, and an anode containing the metal target sealed under high vaccum in a glass envelope. The heated filament in the cathode emits electrons which are accelerated by a high DC voltage and collide with the positive anode target. Two different types of x-ray spectra may be seen. The continuous or bremsstrahlung� spectrum that is always present is produced when the electron Kp Ka I .min . penetrates through the outer electron cloud and is abruptly accelerated by the large positive charge on the nucleus of a heavy atom. The production of x-rays increases with increasing atomic number but is typically no more than 1% efficient, the remaining energy appearing as heat in the target metal. The sharp line spectra that can be seen at higher voltages occur when the incident electrons eject an inner shell electron, such as n = 1 shell electron. The spectral line is produced when an electron, say from n = 2, fills the vacancy in the n = 1 shell, emitting an x-ray photon whose energy corresponds to the energy difference between the n = 2 and n = 1 shells. �The intensity of x-rays is proportional to the number of photons created. The photon energy E = hf = hc/. where h is Planck�s constant and c is the speed of light. Figure 1 is a sketch of intensity versus wavelength for a molybdenum target with an accelerating voltage of 35,000 V. 15. Figure shows that the continuous x-ray spectrum has a minimum (cut-off) wavelength. No shorter wavelengths are emitted from the tube. This minimum wavelength corresponds to : (A) the smallest number of emitted photons (B) the highest energy photons emitted (C) the type of cathode used (D) the type of anode material used. 16. The sharp K. peak in figure 1 corresponds to an electron transition from state n = 2 to n = 1, where K. corresponds to a transition from state n = 3 to n = 1. The K. line peak is higher than the K. because� (A) it is due to a higher atomic number target metal (B) K. is the more energetic transition (C) the K. transition is more probable than the K. transition, so its intensity is higher (D) the K. transition occurs in the valence shells instead of the inner shells 17. The current to the heated filament in the cathode is increased while the accelerating voltage is kept constant. This increased current increases the number of electrons striking the target increasing the overall intensity. What effect does this have on the minimum wavelength value ?

(A) The minimum value will move to shorter wavelength values. (B) It will move to longer wavelength values (C) There will no longer be a cutoff wavelength. (D) The minimum wavelength will remain the same. 18. If the accelerating voltage, V0, increased while keeping the filament current constant, the overall intensity will also increase. What effect will this have on the wavelength position where the two peaks are observed? (A) They will occur at the same wavelengths (B) The peaks occur at shorter wavelengths due to the higher wavelengths due to the higher energies available. (C) They may diappear because all energies may exceed those of the n = 3 to n = 1 transition (D) The K. occur at longer wavelengths but the K. occur at shorter wavelengths. 19. Because the efficiency of x-ray production increases with increasing atomic number, Z, it would seem that lead (Z = 82) would be a better target material than tungsten (Z = 74). However, tungsten (also used in ordinary light bulb filaments) is the most common target meal, whereas lead is never used. this is because- (A) lead has too many electrons (B) lead becomes radioactive (C) lead would melt, whereas tungsten has a very high melting point (D) lead will not get hot enough to produce and x-rays.

MODERN PHYSICS www.physicsashok.in 144 PASSAGE : 5 Student in a medical physics class are given the assignment of planning a nuclear medicine facility. They not only design the rooms and choose the major equipment they also will have to solve elementary problems dealing with treatment, doses, radiation protection and safety, and the general principles of physics of typical isotopes that might be used in diagnostic nuclear medicine. They are required to be familiar with concepts of half life, half-thickness for shielding and the decay schemes of representative isotopes. 20. The most common isotope used in diagnostic work is Technicium. It is furnished from a generator or �cow� in which the negative beta decay of Molybdenum -99 produces the desirablemetastable isotope of Technicium according to the following decay scheme ? 42Mo99 . ZTcA + �1e0 What are the atomic number , Z and mass number A, of the Tc isotope ? (A) 41, 99 (B) 42, 99 (C) 43, 98 (D) 43, 99 21. If the Mo99 has a physical half-life of 67 hours, about what fraction is left after 5.5 days ? (A) 0.10 (B) 0.25 (C) 0.40 (D) 0.45 22. This isotope of Technicium has a physical half life of 6 hours. When it is tagged onto a polyphosphate carrier used for a certain procedure, there is a biological half-life of 12 hours (for the biological excretion of the carrier). What is the �effective half life� in this case ? (A) 4.0 hours (B) 7.5 hours (C) 10 hours (D) 14 hours 23. The �cow� was milked at 8.00 A.M. At 2.00 PM the activity of the milked sample is measured by a technician and found to be 200 millicuries. What was the activity of the Technicium at 8.00 A.M. (A) 100 mCi (B) 150 mCi (C) 300 mCi (D) 400 mCi 24. Thallium -201 is used for myocardial prefusion studies of the heart. It decays by electron capture when the nucleus captures one of the atom�s own orbital electrons (converting a proton in the nucleus into an uncharged neutron), with the emission of gamma rays used for the imaging. What are the atomic number Z, the mass number, A, of the Mercury isotope produced in the decay of the Thallium -201 ? 81Tl201 + 1e0 . ZHgA + Is (A) 80, 201 (B) 80, 202 (C) 81, 201 (D) 81, 203 25. The Thallium -201 half-life is 74 hours. If the sample has an activity of 80 millicuries initially, what will be the activity after 9.25 days ? (A) 2.5 mCi (B) 5 mCi (C) 10 mCi (D) 20 mCi 26. One advantage of the Thallium isotope is the �low whole body absorbed dose per millicurie�, which for Tl-201 is 70 millirads/millicurie. If the recommended amount to be injected for a heart scan is 10 microcurie per kg of body mass, what would be the whole body dose in millirads for a 70 kg person ? (A) 34 mrad (B) 49 mrad (C) 72 mrad (D) 134 mrad 27. Another feature that makes Technicium a desirable isotope for diagnostic nuclear medicine use is that it is a �pure gamma emitter.� What is the meaning of the terminology �pure gamma emitt

er�? (A) The gamma radiation stays in the patient�s tissue while the electrons are detected. (B) Particles emitted cannot escape tissue while the gama radiation escapes (C) The isotope decay emits only penetrating gamma radiation that can escape from tissue and be detected. (D) The isotope decay emits energetic electrons that act like gamma rays. 28. A patient is injected with Technicium and is estimated to have received a whole body dose of 400 millirads. If the �Quality Factor� is 1 for gamma radiation and 3 for low energy neutrons, what was the doese received by the patient in rem units ? (A) 4 mrems (B) 400 mrems (C) 1200 mrems (D) 4000 mrems 29. The half-thickness of lead for the absorption of the gamma radiation from a particular isotope is 0.4 cm of lead. How many half thicknesses are necessary to reduce the radiation penetration to less than 1% and how thick would the lead be ? (A) 2 half - thickness, 1.2 cm (B) 4 half-thicknesses, 3.2 cm (C) 7 half-thicknesses, 2.8 cm (D) 11 half-thicknesses , 4.4 cm PASSAGE : 6 Medical researchers and technicians can track the characteristic radiation patterns emitted by certain inherently unstable isotopes as they spontaneously decay into other elements. The half life of a radioactive isotope is the amount of time necessary for one-half of the initial amount of its nuclei to decay. the decay curves of isotopes 39Y90 and 39Y91 are graphed below as functions of the ratio of N, the number nuclei remaining after a given period to N0, the initial number of nuclei.

MODERN PHYSICS www.physicsashok.in 145 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1 2 3 4 5 6 Time (days) N/N0 39Y90 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 30 60 90 120 150 180 Time (days) N/N0 39Y91 30. The half-life of 39Y90 is approximately : (A) 2.7 days (B) 5.4 days (C) 27 days (D) 48 days 31. What will the approximate ratio of 39Y90 to 39Y91 be after 2.7 days if the initial samples of the two isotopes contain equal numbers of nuclei ? (A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 10 : 1 32. When inhaled by humans, 39Y90 accumulates in the gastrointestinal tract, whereas 39Y91 accumulates in the bones. If the total amount of each isotope inhaled goes to the specified area, which of the following situations will exist three days after a patient inhales these substances, assuming none of the isotopes leave the assuming none of the isotopes leave the specified areas due to physiological factors ? (A) The amount of 39Y91 in the gastrointestinal tract will be approximately equal to the total amount inhaled. (B) The amount of 39Y90 in the bones will be approximately one-half of the total amount inhaled (C) The amount of 39Y90 in the gastrointestinal tract will be approximately one-half of the total amount inhaled (D) None of the 39Y91 inhaled will be left in the bones. 33. Approximately how many 39Y91 nuclei will exist after three half -lives have passed, if there are 1,000 nuclei to begin with ? (A) 50 (B) 125 (C) 250 (D) 500 34. Which of the following conclusions is / are supported by the information given in the passage ?

I. 39Y90 is less stable than 39Y91 II. Only one-quarter of the original amount of 39Y90 will remain after 116 days. III. 39Y90 and 39Y91 are both radioactive (A) I only (B) III only (C) I and II only (D) I and III only PASSAGE : 7 Scientists are working hard to develop nuclear fusion reactor. Nucleiof heavy hydrogen, 21 H , known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D�D reaction is 21 H + 21 H . 32He + n+ energy. In the core of fusion reactor, a gas of heavyhydrogen is fullyionized into deuteron nuclei and electrons. This collection of 21 H nuclei and electrons is known as plasma. The nuclei move randomlyin the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no materialwall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t0 before the particles fly away fromthe core. Ifn is the density(number/volume) ofdeuterons, the product nt0 is calledLawsonnumber. In one of the criteria, a reactor is termed successful ifLawson number is greater than 5 × 1014s/cm3. It may be halpful to use the following : Boltzmann constant k = 8.6 × 10�5 eV/K; 2 0 e 4.. = 1.44 × 10�9 eVm. [JEE, 09]

MODERN PHYSICS MODERN PHYSICS 35. In the core of nuclear fusion reactor, the gas becomes plasma because of (A) strongnuclearforceactingbetweenthedeuterons (B) Coulomb force acting between the deuterons (C) Coulomb force acting between deuteron-electron pairs (D) the high temperature maintained inside the reactor core. 36. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each withkinetic energy1.5 kT, whenthe separationbetweenthemis large enoughto neglect Coulomb potential energy.Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10�15 m is in the range. (A) 1.0 × 109 K < T < 2.0 × 109 K (B) 2.0 × 109 K < T < 3.0 × 109 K (C) 3.0 × 109 K < T < 4.0 × 109 K (D) 4.0 × 109 K < T < 5.0 × 109 K 37. Results of calculations for four different designs of a fusion reactor using D�D reaction are given below. Which of these is most promising based on Lawson criterian ? (A) deuteron density = 2.0 × 1012 cm�3, confinement time = 5.0 × 10�3 s (B) deuteron density = 8.0 × 1014 cm�3, confinement time = 9.0 × 10�1 s (C) deuteron density = 4.0 × 1023 cm�3, confinement time = 1.0 × 10�11 s (D) deuteron density = 1.0 × 1024 cm�3, confinement time = 4.0 × 10�12 s www.physicsashok.in 146

MODERN PHYSICS www.physicsashok.in 147 Level � 2 1. When a surface is irridated with light of . = 4950A., a photo current appears which vanishes if a retarding potential greater then 0.6V is applied across the photo tube. When a different source of light is used, it is found that the critical retarding potential is changed to 1.1 V. what is the work function of the surface and the wavelength of the second source ? If the photo-electrons (after emission from the source) are subjected to a magnetic field of 10 tesla what changes will be observed in the above two retarding potentials? 2. A particle of charge equal to that of an electron and mass 208 times the mass of the electron (called a mu-meson) moves in a circular orbit around a nucleus of charge +3e. (Take the mass of the nucleus to be infinite.) Assuming the Bohr model of the atom to be applicable to this system, (i) derive an expression for the radius of the nth Bohr-orbit, (ii) find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom, and (iii) find the wavelength of the radiation emitted when the .�meson jumps from the third orbit to the first orbit. (Rydberg�s constant = 1.097 x 107 m�1 ) 3. Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975A.. How many different lines are possible in the resulting spectrum ? Calculate the longest wavelength amongst them. You may assume the ionization energy for hydrogen atom as 13.6 eV. 4. An X-ray tube works at a potential difference of 100,000 V. Only 0.1% of the energy of cathode rays is converted into X-ray radiation and heat is generated in the target at the rate of 120 calorie per second. What current does the tube pass and what is the energy and velocity of an electron when it reaches the target? 5. A laser emits a light pulse of duration . . 0.13 ms and energy E = 10 J. Find the mean pressure exerted by such a light pulse when it is focussed into a spot of diameter d = 10 .m on a surface perpendicular to the beam and possessing a reflection coefficient . = 0.5 6. A short light pulse of energy E = 7.5 J falls in the form of a narrow and almost parallel beam on a mirror plate whose reflection coefficient is . = 0.60. The angle of incidence is 30°. In terms of the corpuscular theory find the momentum transferred to the plate. 7. The binding energy of an electron in the ground state of He atom is equal to E0 = 24.6 eV. Find the energy required to remove both electrons from the atom. 8. A stationary He+ ion emitted a photon corresponding to the first line of the Lyman series. That photon liberated a photoelectron from a stationary hydrogen atom in the ground state. Find the velocity of the photoelectron.

9. Taking into account the motion of the nucleus of a hydrogen atom, find the expressions for the electron�s binding energy in the ground state and for the Rydberg constant. How much (in per cent) do the binding energy and the Rydberg constant, obtained without taking into account the motion of the nucleus, differ from the more accurate corresponding values of these quantities? 10. Calculate the separation between the particles of a system in the ground state, the corresponding binding energy, and the wavelength of the first line of the Lyman series, if such a system is (a) a mesonic hydrogen atom whose nucleus is a proton ( in a mesonic atom an electron is replaced by a meson whose charge is the same and mass is 207 that of an electron); (b) a positronium consisting of an electron and a positron revolving around their common centre of masses.

MODERN PHYSICS www.physicsashok.in 148 11. In accordance with classical electrodynamics an electron moving with an acceleration a loses its energy due to radiation as : 2 32 a 3c2e dt d . . . Estimate the time during which an electron moving in a hydrogen atom along a circular orbit of radius r = 50 pm would have fallen onto the nucleus. Assume a to be directed permanently towards the nucleus. 12. The Earth revolves round the Sun due to gravitational attraction. Suppose that the Sun and the Earth are point particles with their existing masses and that Bohr�s quantization rule for angular momentum is valid in the case of gravitation. (a) Calculate the minimum radius the Earth can have for its orbit. (b) What is the value of the principal quantum number n for the present radius? Mass of the Earth = 6.0 x 1024 kg, mass of the Sun = 2.0 x 1030 kg, Earth-Sun distance = 1.5 x 1011 m. 13. A parallel electron beam is obtained from the application of accelerating voltage V0 = 2 x 104 V. These electrons are sent travelling in the direction normal to an infinitely long straight copper wire of radius r0 = 10�8 m as shown in figure. The wire carries uniform positive charge with charge density . = 4.4 x 10�11 c/m, the distance of the electrons closest b L Electron Beam Screen approach to the wire if uncharged is represented by bmax = 10�4 m. The electrons after passing the charged copper wire land on the screen located at distance L = 0.3 cm ( L >.b) from the wire. At the beginning of the experiment the electron beam is confined within the normal distance to the wire of b ( collision parameter) (see Figure). (a) Determine electric field E due to the charged copper wire and sketch a graph of electric field E as a function of distance measured from the axis of the wire to inside as well as outside the wire. (b) Calculate the angle of deflection of electron beam which has the collision parameter b and the electron does not collide with the wire. If f . is a small angle between the direction of the original velocity of the electron and the direction of final velocity of the electron arriving at the screen. Calculate f . . (c) Show that two plane waves representing deflected beams in the upper and lower parts give interference pattern on the screen. Calculate the number of bright bands in the interfence pattern. 14. To stop the flow of photoelectrons produced by electromagnetic radiation incident on a certain metal, a

negative potential of 300 volts is required. If the photoelectric threshold of the metal is 1500A., what is the frequency of the incident radiation? Is this radiation visible? 15. A toy truck has dimensions as shown in Figure and its width normal to the plane of this paper is d. Sun rays are incident on it as shown in figure. If intensity of sun rays is . and all surfaces of truck are perfectly black, calculate tension in thread used to keep truck stationary. Neglect fraction. . Thread b a h 16. Amonochromatic beam of light ( . = 4900 A°) incident normally upon a surface produces a pressure of 5 x 10�7 N/m2 on it. Assuming that 25% of the light incident is reflected and the rest absorbed, find the number of photons falling per second on a unit area of thin surface. 17. A beam of light consists of four wavelength 4000A., 4800A., 6000A.and 7000A., each of intensity 1.5 x 10�3 wm�2. The beam falls normally on an area 10�4 m2 of a clean metallic surface of work function 1.9 eV.Assuming no loss of light energy calculate the number of photo electrons liberated per second. 18. A single electron orbits around a stationay nucleus of charge +Ze, where Z is a constant and e is the magnitude of the electric charge. It requires 47.2 eV to excite the electronfromthe second Bohrorbit tothe third Bohr orbit. Find :

MODERN PHYSICS www.physicsashok.in 149 (i) the value of Z (ii) the energy required to excite the electron from the third to the fourth Bohr orbit. (iii) The wavelength of the electromagnetic radiation required tomove the electron from. (iv) Thekinetic energy, the potentialenergy and the angular momentumof the electron in the first Bohr ortib. (v) the radius of the first Bohr orbit. 19. Radiations from hydrogenic C gas corresponding to 2nd line of Lyman series falls on a hydrogenic atom where rotating particle�s mass and charge are unknown. Nucleus contains one proton only. This atom is excited to 4th excited energy level. (a) Find a relation for valid values of mass and charge of the rotating particles. (b) Find the maximum K.E.of the electron ejected due to aforesaid radiation falling on a hydrogenic potassium having some of the atoms in the ground energy level & some in the 4th energy level. 20. Stopping potentials of 24, 100, 110 and 115 kV are measured for photoelectrons emitted from a certain element when it is radiated with monochromatic x-ray. If this element is used as a target in an x-ray tube, what will be the wavelength of K line? 21. Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photo-electrons emitted form sodius is 0.73 eV. The work fraction for sodium is 1.82 eV. Find : (i) the energy of the photons causing the photoelectric emission. (ii) the quantum numbers of the two levels involved in the emission of these photons. (iii) the charge in the angular momentum of the eletron in the hydrogen atom in the above transition. (iv) the recoil speed of the emitting atom assuming it to be at rest before the transition. (Ionization potential of hydrogen is 13.6 eV). 22. A gas of hydrogen like atoms can absorb radiations of 68 eV. Consequently, the atoms emit radiations of only three different wavelengths. All the wavelengths are equal or smaller than that of the absorbed photon. (a) Determine the initial state of the gas atoms. (b) Identify the gas atoms. (c) Find the minimum wavelength of the emitted radiation. (d) Find the ionization energy and the respective wavelength for the gas atoms. 23. According to the Thomson model, a helium atom consists of a cloud of positive charge, within which two electrons sit at equlibrium positions. Assume that the positive cloud has a charge +2e uniformly distributed over the volume of a sphere of radius 0.50 A.. (a) Find the equilibrium position of the two electrons. Assume that the electrons are symmetrically placed with respect to the centre. (b) What is the frequency of small radial oscillations of the electrons about th

eir equilibrium positions. Assume that the electrons move symmetrically with identical amplitudes. 24. The Rydberg constants for hydrogen and singly ionized helium are R1 and R2 respectively. Find the ratio of the mass of the proton to that of the electron in terms of R1, R2 and R. . 25. A sample contains 10�2 kg each of two substances A and B with half lives 4 second and 8 second respectively. Their weights are in the ratio of 1.2. Find the amounts of A and B after an interval of 16 second. 26. A sample of uranium is a mixture of three isotopes 92U234, 92U235 and 92U238 present in the ratio of 0.006%, 0.71% and 99.284% respectively. The half-lives of these isotopes are 2.5 x 105 years, 7.1 x 108 years and 4.5 x 109 years respectively. Calculate the contribution to activity (in %) of each isotope in this sample.

MODERN PHYSICS www.physicsashok.in 150 27. Polonium ( 84Po210 ) emits 2He4 particles and is converted into lead (82Pb206). This reaction is used for producing electric power in a space mission. Po210 has half life of 138.6 days. Assuming an efficiency of 10% for the thermoelectric machine, howmuch Po210 is required to produce 1.2 x 107 J of electric energy per day at the end of 693 days. Also find the initial activity of the material. (Given : masses of nuclei Po210 = 209.98264 a.m.u., Pb206 = 205.97440 a.m.u., 2He4 = 4.00260 a.m.u., 1 a.m.u. = 931 MeV and Avogadro number = 6x1023/mol. 28. 10�3 kg of radioactive isotope (atomic mass 226) emits 3.72 x 1010 . -particles in a second. Calculate the half-life of the isotope. If 4.2 x 102 joule is released in one hour in this process, what is the average energy of the . -particle? 29. In an ore containing uranium, the ratio of U238 to Pb206 nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of U238. Take the half-life of U238 to be 4.5 x 109 year. 30. A7 Li target is bombarded with a proton beam current of 10�4 A for 1 hour to produce 7Be of activity 1.8 x 108 disintegrations per second. Assuming that one 7Be radioactive nuclei is produced by bombarding 1000 protons, determine its half-life. Level � 3 1. Asmall quantityof solution containing 24Na radionuclide (half life 15 hours) or activity1.0microcurie is injected into the blood of a person.Asample of the blood of value 1 cm3 taken after 5 hours shows an activity of 296 disintegrations per minute. Determine the total value of blood in the bodyof the person. Assume that the radioactive solutionmixes uniformly in the blood of the person. (1 Curie = 3.7 × 1010 disintegrations per second) [JEE, 94] 2. At a given instant there are 25%undecayed radio-active nuclei in a sample.After 10 sec the number of undecayed nuclei remains to 12.5%. Calculate : [JEE, 96] (i)mean-life ofthe nuclei and (ii)The time inwhich the number ofundecayed nuclearwill further reduce to 6.25%ofthe reduced number. 3. Consider the following reaction ; 2H1 + 2H1 = 4He2 + Q. [JEE, 96] Mass of the deuteriumatom= 2.0141 u;Mass of the heliumatom= 4.0024 u This is a nuclear _______ reaction inwhich the energyQis released is _______MeV. 4. The element Curium 248 96 Cmhas a mean life of 1013 seconds. Its primary decaymodes are spontaneous fission and . decay, the formerwith a probabilityof 8%and the laterwitha probabilityof92%. Eachfission releases 200MeVof energy.Themasses involved in . decayare as follows : 248 96 Cm= 248.072220 u, 244 94 Pu = 244.064100 u & 42He = 4.002603 u. Calculate the power output froma sample of 1020 Cmatoms. (Iu = 931MeV/C2) [JEE, 97]

5. Nuclei of a radioactive elementAare being produced at a constant rate .. the element has a decay constant ..At time t = 0, there areN0 nucleiof the element. [JEE, 98] (a) Calculate the number Nof nuclei ofAat time t. (b) If . = 2N0., calculate the number of nucleiofAafter one halflife ofA&also the limiting value ofNas t ....

MODERN PHYSICS www.physicsashok.in 151 6. Photoelectrons are emittedwhen 400 nmradiationis incident on a surface ofwork function 1.9 eV. These photoelectrons pass througha regioncontaining .-particles.Amaximumenergyelectron combineswith an .-particle to formaHe+ ion, emittinga single photoninthis process.He+ ions thus formed are intheir fourth excited state. Find the energies in eVofthe photons, lying in the 2 to 4eVrange, that are likelyto be emitted during and after the combination. [Take, h = 4.14 × 10�15 eV�s] [JEE, 99] 7(a). Ahydrogen-like atomof atomic number Z is in an excited state of quantumnumber 2 n. It can emit a maximumenergyphoton of 204 eV. If itmakes a transition to quantumstate n, a photon ofenergy40.8 eV is emitted. Find n, Z and the ground state energy(in eV) for this atom.Also, calculate theminimumenergy (in eV) that can be emitted bythis atomduring de-excitation. Ground state energy of hydrogen atomis � 13.6 eV. [JEE, 2000] (b). When a beamof 10.6 eV photon of intensity 2W/m2 falls on a platinumsurface of area 1 × 104 m2 and work function5.6 eV, 0.53%ofthe incident photons eject photoelectrons. Find the number ofphotoelectrons emitted per sec and theirminimumandmaximumenergies in eV. [JEE, 2000] 8. Ahydrogenlike atom(described bytheBohrmodel) is observed to emit sixwavelengths, originating from all possible transitionbetween a group of levels. These levels have energies between � 0.85 eVand �0.544 eV(including both these values) [JEE, 02] (a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in these transitions. 9. Two metallic platesAand B each of area 5 × 10�4m2, are placed at a separation of 1 cm. Plate B carries a positive charge of 33.7 × 10�12 C.Amonochromatic beamof light,with photons of energy 5 eVeach, starts falling onplateAat t = 0 so that 1016 photons fall on it per squaremeter per second. assume that one photoelectron is emitted for every106 incident photons.Also assume that all the emitted photoelectrons are collected by plateB and thework dunction of plateAremains constant at the value 2 eV. Determine (a) the number of photoelectrons emitted up to t = 10 sec. (b) themagnitude of the electric field between the platesAand B at t = 10 s and (c) the kinetic energy of the most energetic photoelectron emitted at t = 10 s when it reaches plate B. (Neglect the time taken by photoelectron to reach plate B) [JEE, 02] 10. Frequencyof a photon emitted due to transition ofelectronofa certain elemrnt fromLtoKshellis found to be 4.2 × 1018 Hz. UsingMoseley�s law, find the atomic number of the element, given that the Rydberg�s constant R = 1.1 × 107 m�1. [JEE, 03] 11. In a photoelectric experiment set up, photons of energy 5 eVfalls on the cathode havingwork function 3 eV. (a) if the saturation current is iA = 4µA for intensity10�5W/m2, then plot a graph between anode potential and current. (b)Also drawa graph for intensity of incident ratiation of 2 × 10�5W/m2 ? [JEE, 03]

12. Aradioactive samle emits n .-particles in 2 sec. In next 2 sec it emits 0.75 n .-particles, what is themean life of the sample ? [JEE, 03]

MODERN PHYSICS www.physicsashok.in 152 13. The potential energy of a particle ofmassmis given by 0 E 0 x 1 V(x) 0 x 1 . . . . . . . . . . .1 and .2 are are the de-Brogliewavelengths of the particle, when 0 . x . 1 and x > 1 respectively. If the total energy of particle is 2E0, find .1/.2. [JEE, 05] 14. Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of heliumnucleus is (14)1/3. Find [JEE, 05] (a) atomic number of the nucleus (b) the frequency ofK. line of the X-ray produced. (R = 1.1 × 107 m�1 and c = 3 × 108 m/s) 15. Inhydrogen-like atom(z=11), nth line ofLymanseries haswavelength. equalto thede-Broglie�swavelength of electron in the level fromwhich it originated.What is the value of n ? [Take :Bohr radius (r0) = 0.53Å and Rydberg constant (R) = 1.1 × 107m�1] [JEE, 06] Answer Key Assertion-Reason Q. 1 2 3 4 5 6 7 8 9 10 Ans. B C D B A A B A A B Match the Column 1. (A) . (U), (B) . (R), (C) . (P), (D) . (S) 2. (A) . (S), (B) . (T), (C) . (Q), (D) . (P) 3. (A) . (R), (B) . (P), (C) . (T), (D) . (U) 4. (A) . (R), (B) . (U), (C) . (V), (D) . (Q) 5. (A) . (Q), (B) . (R), (C) . (S), (D) . (P) 6. (A) . (PQ), (B) . (PR), (C) . (PS), (D) . (PQR) 7. (A) . (P), (B) . (T), (C) . (U), (D) . (R) 8. (A) . (PR), (B) . (QS), (C) . (P), (D) . (Q) 9. (A) . (PQT), (B) . (Q), (C) . (S), (D) . (S)

MODERN PHYSICS www.physicsashok.in 153 Level � 1 Q. 1 2 3 4 5 6 7 8 9 10 Ans. D C D B C A C D D A Q. 11 12 13 14 15 16 17 18 19 20 Ans. B B B D A D A C ABC C Q. 21 22 23 24 25 26 27 28 29 30 Ans. C D D C D C B D D B Q. 31 32 33 34 35 36 37 38 39 40 Ans. B C AD BC ABC BD ABD AB AC BD Q. 41 42 43 44 45 46 47 48 49 50 Ans. ABC CD AC AB AB AC AD AC AB C Q. 52 53 54 Ans. AD C B Q. 57 59 60 61 62 63 64 65 Ans. C A B C AC A B A Q. 66 67 68 69 70 71 72 Ans. B BD C C A A A 56 (a-C), (b-B), (c-B), (d-E), (e-C) 58 (a-C), (b-A) 51 (i-B), (ii-A) (i-CD), (ii-D) 55 Passage Q. 1 2 3 4 5 6 7 8 9 10 Ans. C B A D B B D A D A Q. 11 12 13 14 15 16 17 18 19 20 Ans. B B C A B C D A C D Q. 21 22 23 24 25 26 27 28 29 30 Ans. B A D A C B C B C A Q. 31 32 33 34 35 36 37 Ans. B C B D D A B Level � 2 1. 1.9 eV 0 . . , . = 4125A.2. (i) 8.426 x 10�14 n2 (ii) 25 (iii) 5.478 x 10�11 m 3. n = 4; . = 18800A.4. 50.6 .A 5. (p) . 4.1. .. . .d2c. . 50 atm 6. P . .E/c. 1. .2 . 2. cos 2. . 35 nN.s 7. 4 R 79 eV. 0 . . . . . . 8. .. 2 .R/m.3.1.106 m/s, where m is the mass of the electron. 9. e4 / 2 3 , b . . . . where . is the reduced mass of the system. If the motion of the nucleus is not taken into account, these values (in the case of a hydrogen atom) are greater by m/M . 0.055%, where m and M are the masses of an electron and a proton.

MODERN PHYSICS www.physicsashok.in 154 10. (a) 0.285 pm, 2.53 keV, 0.65 nm; (b) 106 pm, 6.8 eV, 0.243 . m. 11. ps 13 e 4 / r c m t 2 3 2 2e. . 12. (a) 2.3 x 10�138 m (b) 2.5 x 1074 13. (a) 2 r E q L0 .. . . for r > r0 = 0; (b) 0 0 f 4 V q ... . . ; (c) 500 14. v = 7.45 x 1016 cycles/s, no 15. . . . . . .. . . b cos acos hsin sin cd 16. 2.93 x 1020 17. 11.72 x 1011 18. (i) x = 5 ; (ii) 16.53 eV (iii) 36.36 A ; (iv) 1.056 x 10�34 kg m2 s�1 (v) 1.06 x 10�11 m 19. (a) 1.6 x 10�17 2 2 0 . h (b) The radiation connot eject electron. 20. 0.163 A 21. 2.55 eV, electron jumps from 4th to 2nd orbits, 2.11 x 10�34 Js 0.8144 m/s 22. (a) n = 2 (b) z = 6 (c) 28.5 A. (d) 25.32 A. 23. (a) 0.25 A (b) 1.76 x 1016 Hz 24. .. . .. . . . 1 R2 1 RR 10.75 25. NA = 6.25 x 10�4 kg, NB = 2.5 x 10�3 26. 51.41%, 2.13%, 46.45% 27. A0 = 4.57 x 1021 per day, 320 gm 28. T½ = 1573 year, Energy per . particle = 19.5 MeV 29. 1.868 x 109 year 30. T½ = 8.63 x 106 s. Level � 3 1. 6 litre 2. (i) t1/2 = 10 sec., tmeans = 14.43 s; (ii) 40 seconds 3. Fusion, 24 4. . 33.298 µW 5. (a) t t 0 N . 1 [.(1. e.. ) . .N e.. ] . ; (b) 0 0 3N , 2N 2 6. during combination = 3.365 eV; after combination = 3.88 eV(5 . 3) &2.63 eV(4 . 3) 7.(a) n = 2, z = 4; G.S.E.� 217.6 eV;Min. energy = 10.58eV; (b) 6.25×1019per sec, 0, 5 eV 8. (a) 3; (b) 4052.3 nm 9. (a) 5 × 107, (b) 2000 N/C, (c) 23 eV

10. z = 42 11. 8µA I 4µA �2V VP I=2×10�5W/m2 I = 10�5W/m2 12. 1.75n = N0(1�e�4.), 6.95 sec, 2 ln(4 / 3) 13. 2 14. v = 1.546 × 1018 Hz 15. n = 24 �X�X�X�X�

NUCLEAR PHYSICS

DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE, TIRUPATI. PH NO. 9440025125

NUCLEAR PHYSICS www.physicsashok.in 1 RADIO ACTIVITY INTRODUCTION OF ATOMIC NUCLEUS (a) The atomic nucleus consists oftwo types of elementaryparticles, viz. protons and neutrons.These particles are collectivelycalled nucleons. (b) The electrons surround this nucleus to formthe atom. (c) This structure ofatomwas revealed by the experiments ofRutherford inwhich a beamof .-particleswere made to strike a thin gold foil. (d) Nucleus can be regarded as a small spherical volume situated at the centre of atom. (e) Most of themass of the atomis concentrated in the nucleus. (f) Nucleus has a positive charge. PROPERTIES OF AN ATOMIC NUCLEUS Composition: (a) All nucleiare composed of two types of particles protons and neutrons. (b) The onlyexceptionis the ordinary hydrogen nucleuswhich is just a single proton. (c) A proton has a mass of 1.6726 × 10�27 kg and charge +e (= 1.6 × 10�19 C). (d) Aneutron has amass of 1.6750 × 10�27 kg and is electrically neutral. The atomic number Z: This is equal to the number of protons in the nucleus. The neutron numberN: This is equal to the number of neutrons in the nucleus. Themass numberA: This is equal to the number of nucleons (protons + neutrons) in the nucleus. Thus,A= Z + N Symbolically a nucleusXshall be represented as A ZX . Types of Nuclei : Nuclei are of three types: (i) Isotopes: Nuclei having the same Z value but different NandAvalues are called isotopes of the element. eg. 1H1, 1H2 and 1H3 are isotopes of hydrogen. (ii) Isobars: Nuclides (nucleiof different elements) having the sameAvalue but different Z andNvalues are called isobars. e.g. 1H3 and 2He3 are isobars. (iii) Isotones: Nuclides having the same Nvalue but differentAand Z values are known as isotones. e.g. 1H3 and 2He4 are isotones. Radius of atomic nucleus: The size of the nucleus is of the order of fermi (fm). 1 fm = 10�15. Most nucleus are almost sphericalwith an average radius Rgiven by R = R0A1/3 whereAis the mass number and R0 is a constant. R0 ~. 1.2 fm = 1.2 × 10�15 m

NUCLEAR PHYSICS www.physicsashok.in 2 Mass : (a) Nuclearmasses have been accurately determinedwith the help of themass spectrometer. (b) It is expressed in a.m.u. i.e. atomicmass unit. (c) This unit is such that mass of the carbon isotope 6C12 is exactly 12 amu. (d) 1 a.m.u. = 1.66056 × 10�27 kg (e) Matter can be viewed as a condense formof energy. (f) The energy corresponding to themass of a particlewhen it is at rest is called rest mass energy. (g) Rest mass energyis E =mc2, fromEinstein relation. (h) Energy equivalent of 1 a.m.u. is equal to 931.5MeV. i.e. 1 a.m.u. = 931.5MeV/c2. Example 51: (a) Calculate the value of 1 a.m.u. fromAvogadro�s number. (b) Determine the energy equivalent of 1 a.m.u. Sol: (a) Onemole of C12 has a mass of 12 g and containsAvogadro number, NA of atoms. Bydefinition eachC12 atomhas a mass of 12 a.m.u. Thus, 12 g corresponds to (12NA) a.m.u.whichmeans, 1 a.m.u. = A 1g N = 3 23 10 6.022055 10 . . kg or 1 a.m.u. = 1.66056 × 10�27 kg (b) FromEinstein relation, restmass energy E = mc2 Hence, energy equivalent of 1 a.m.u., E = (1 a.m.u.) × c2 = (1.66056 × 10�27 kg) × (3 × 108 m/s)2 = 1.4924 × 10�10 J Since, 1 eV = 1.6 × 10�19 J, . E = 931.5MeV Hence, 1 a.m.u = 931.5MeV/c2 Density: Mass of a nucleus can be taken approximatelyAm, wheremis the mass of proton or a neutron andAis mass number . Mass =Am Also, assuming the nucleus to be a sphere of radius R, its volume is V = 43 .R3 = 30 4 R A 3 . , [. R = R0A1/3] The nuclear densityis thus given by, . = mass volume = 30 Am

4 R A 3 . = 30 3m 4.R It is thus independent of themass numberAand is therefore nearly the same for allnuclei. Putting R0 = 1.2 fm = 1.2 × 10�15mand m= 1.67 × 10�27 kg, we get, . = 2.3 × 1017 kg/m3. This is almost 1014 times the densityofwater. (a) Nuclear density . = 2.3 × 1017 kg/m3 (b) It is nearly the same for all nuclei.

NUCLEAR PHYSICS www.physicsashok.in 3 C48: Find themass density of the oxygen nucleus 8O16. Sol: Volume = 43 .R3 = 30 4 R A 3 . [. R = R0A1/3] = 43 .(1.2)3 × 16 × 10�45 m3, [. R0 = 1.2 × 10�15, A= 16] = 1.16 × 10�43 m3 Mass of oxygen atoms (A= 16) is approximately 16 a.m.u. Therefore densityis . = mass volume = 43 3 16 amu 1.16.10. m = 27 43 3 16 1.66 10 1.16 10 m. . . . . kg/m3 = 2.3 × 1017 kg/m3 NUCLEAR STABILITY (a) High densityof the nucleus suggests a very tight packing of protons and neutrons in it. (b) The Coulomb�s repulsive force between two protons in a nucleus is about 1036 times as large as the gravitational force between them. It is therefore surprising that a nucleus should be so stable. (c) Nuclei are stable because of the presence of another force, called the nuclear force. Nuclear force : It arises due to interaction betweenprotons, protonwith neutrons, and neutronwith neutrons.This force is essentially a very strong attractive force and overcomes the electrostatic repulsion between the proton inside the nucleus. Properties of Nuclear forces : (a) These are strong attractive forces. (b) These are about 100 times stronger than the Coulomb�s force. (c) These are short ranged forces (effective upto 10 fm). (d) They contain a small component of repulsive force which is effective up to a distance of the order of 0.5 fmor 0.5 × 10�15mor less. This repulsive component prevents the collapse of the nucleus. (e) These forces are charge independent. (f) Let d be the range of the effectiveness of nuclear forces, then 0.5 fm . d . 10 fm (g) Let Fpp, Fpn and Fnn denote themagnitude of the nuclear force by a proton on

a proton, by a proton on a neutron and bya neutron ona neutron respectively. Then for a separation of 1 fm, Fpp = Fpn = Fnn. N/Z ratio : (a) N/Zratio inside a nucleus is responsible for stability of a nucleus. N = Z line N v/s Z curve N (b) It must be greater than or equal to unity and less than 1.6 Z i.e. 1 . NZ . 1.6.

NUCLEAR PHYSICS www.physicsashok.in 4 (c) Reason forN/Zratio to be greater than unityis due to the fact that protons are positivelycharged and repel on another electrically.This repulsion becomes so great in nucleiwithmore than 10 protons or so that an excess of neutrons, which produce onlyattractive nuclear forces, is required for stability. ThusN/Z ratio increaseswith increase in Z. (d) When excess of neutrons or protons in a nuclide is there then the nuclide .-decayor .-decayto achieve the requiredN/Zratio for stability.This causes radioactive disintegrations of nuclides. Binding Energy (B.E.) : The binding energy is equal to thework thatmust be done to split the nucleus into particles constituting it. Hence, Energyof nucleus+B.E. = Energyofeach nucleon individually. Let Mass of nucleus =M, Mass of neutron =mn, and, Mass of proton = mp . Rest mass energyof nucleus =Mc2, . Rest mass energyof neutron =mnc2, . Rest mass energyof proton =mpc2. Thus, Mc2 + B.E. = Z mpc2 + (A � Z)mn c2 . B.E. = [M� {Zmp + (A� Z)mn}]c2 The quantity,M� {Zmp + (A� Z)mn} is called mass-defect (.m) . Mass � defect = .m=M� {Z mp + (A � Z)mn} . B.E. = .m.c2 If .mis in a.m.u., then B.E. = .m× 931.5MeV or B.E. = [M� {Z mp + (A � Z)mn}] × 931.5MeV NOTE: Negative sign of B.E. represents boundedness of nucleons inside the nucleus. C49: Find the binding energy of 56 26Fe .Atomicmass of 56Fe is 55.9349 u and that of 1His 1.00783 u.Mass of neutron = 1.00867 u. Sol: Z = The number of protons in 56 26Fe = 26 and the number of neutrons,A� Z = 56 � 26 = 30 Then binding energy of 56 26Fe = [M� {Zmp + (A� Z))mn}]c2 = � [26 × 1.00783 u + 30 × 1.00867 u � 55.9349 u]c2 = � (0.52878 u)c2 = � (0.52878 u) (931MeV/u) = � 492 MeV Negative sign indicates boundedness of nucleons. . B.E. = 492MeVinmagnitude. Binding energy per nucleon : Binding energyper nucleonis obtainedbydividing the binding energyof the nucleus bythe numberA of nucleons in the nucleus. i.e. B.E. per nucleon = B.E. A 0 50 100 150 200 250 246 8 10

56 26Fe 4He 6Li Binding energy per nucleon, MeV Mass number, A

NUCLEAR PHYSICS www.physicsashok.in 5 (a) The adjacent figure shows the dependence of the B.E. per nucleon, B.E./Aonthemass number Aofthe nucleus. (b) Nucleons in nucleiwithmass number from50 to 60 have the highest B.E. TheB.E./Afor these nuclei amounts to 8.7MeVand gradually decreaseswith increasingA. (c) B.E. per nucleon is highest for 56 26Fe . Example 52: Find the binding energy of 126C ?Also find the binding energy per nucleon. Mass of 6C12 atom = 12 a.m.u. Mass of proton = 1.00759 a.m.u. Mass of neutron = 1.00898 a.m.u. Sol: M = mass of C12 atom = 12 a.m.u. mp = mass of proton = 1.00759 a.m.u. mn = mass of neutron = 1.00898 a.m.u. Z = number of proton = 6 A� Z = number of neutrons = 12 � 6 = 6 . Mass-defect, .m = M� {Zmp + (A� Z)mn} = 12 � (6 × 1.00759 + 6 × 1 × 1.00898) in a.m.u. = (12 � 12.009) a.m.u = �0.099 a.m.u. . B.E. = .m× 931.5MeV = �0.099 × 931.5 MeV = �92.22MeV Hence, Binding energy= �92.22MeV (Negative sign indicates boundedness of the nucleon) Binding energy per nucleon = B.E. A = 92.22 12 MeV= 7.68MeV NUCLEAR COLLISIONS (a) Anuclear reaction in which a collision between particle a and nucleus X producesY and particle b is represented as a + X Y+ b (b) The reaction is sometimes expressed inthe shorthand notationX(a, b)Y. (c) Reaction are subjected to the restrictions imposed by (i) The conservationof charge, (ii) The conservationof energy, (iii) The conservationofmomentum, and (iv) The conservationof angularmomentum. Q-Value or Energy of a reaction : Let m2,m3 are nuclearmasses ofXandYrespectively. a m1 K1 X m2 K2 Before collision . Y m3 K3 X m4 K4 After collision Initial energy: Ei = m1c2 + m2c2 + K1 + K2 Final energy: Ef = m3c2 + m4c2 + K3 + K4

NUCLEAR PHYSICS www.physicsashok.in 6 Since, Ei = Ef, (fromenergyconservation) . [(m1 + m2) � (m3 + m4)]c2 = (K3 + K4) � (K1 + K2) The energy, [(m1 + m2) � (m3 +m4)]c2, that is released or absorbed in a nuclear reaction is called the QValue or disintegration energyof the reaction. Hence, Q = [(m1 + m2) � (m3 + m4)]c2 J or, Q = [(m1 + m2) � (m3 + m4)] 931.5MeV, when masses are in a.m.u. Mass defect: The quantity [(m1 +m2) � (m3 +m4)] is called themass defect of the reaction and is given by .m= (m1 + m2) � (m3 + m4) ina.m.u. Q = .m. 931.5 MeV Example 53:Aneutron breaks into a proton and electron. Calculate the energy produced in this reaction in MeV. Mass of an electron = 9 × 10�31 Kg, Mass of proton = 1.6725 × 10�27 kg, Mass of neutron = 1.6747 × 10�27 Kg. Speed of light = 3 × 108m/s. Sol: 0n1 1H1 + �1e0 Mass defect, .m = [mass of neutron � (mass of proton +mass of �1e0)] = 1.6747 × 10�27 kg � (1.6725 + 0.0009) × 10�27 kg = 0.0013 × 10�27 kg . Energyreleased, Q = .m.c2 = 1.3 × 10�30 × (3 × 108)2 kg �m2/s2 = 1.17 × 10�13 J = 13 19 1.17 10 1.6 10 . . . . eV = 0.73MeV Example 54. Find theQ-value of the reaction 1H2 + 3Li6 . 3Li7 + 1H1 The rest masses of 1H2, 3Li6.. 3Li7, and 1H1are, respectively, 2.01410 amu, 6.01513 amu, 7.01601 amu and 1.00783 amu. Sol. Suppose 1H2 + 3Li6 . 3Li7 + 1H1 + Q. Totalmass of left-hand side = 2.01410 + 6.01513 = 8.02933 amu Totalmass on right-hand side = 7.01601 + 1.00783 = 8.02384 amu . 8.02933 = 8.02384 + Q or Q = 0.00549 amu = 0.00549 × 931 MeV (. 1 amu = 931MeV) Q = 5.1 MeV INTRODUCTION OF RADIOACTIVITY (a) In 1896, Becquerel discovered accidentally that uraniumsalt crystals emit an invisible radiationwhich affected a photographic plate even though it was properly covered. (b) In 1898,Marie and PierreCurie and otherworkers showed thatmanyother substances also emitted similar radiations.

NUCLEAR PHYSICS www.physicsashok.in 7 (c) The propertyof spontaneous emission ofradiation fromthe substance is called radioactivityand such type of substance is called radioactive substance. (d) Radioactivityis due to the decay or disintegrationof unstable nuclei. (e) The radiations arebeing emitted fromthenucleihence it is anuclear phenomenon, not anatomicphenomenon. (f) Some examples of radioactive substances are:U, Ra, Th, Po and Np. (g) Electronic configuration ofatomdoes not have anyrelationshipwith radioactivity. (h) Radioactivity is explained on the basis of quantummechanics. (i) No single phenomenon has played so significant role in the development of nuclear physics as radioactivity. (j) It is not influenced byexternal parameters such as pressure, temperature, chemical reaction (combination) or phase ofmatter. C50:Uraniumsalt crystals emit (a) visible radiation (b) invisible radiation (c) anytype of electromagnetic radiation (d) soundwaves Sol: This is according to discovery ofradioactivityofBecquerel. C51: The radioactivityis a/an (a) optical phenomenon (b)Atomic phenomenon (c) nuclear phenomenon (d) photoelectric phenomenon Sol: Nuclear radiations are obtained fromthe nuclei hence it is a nuclear phenomenon. RADIOACTIVE DECAY (a) The decayof radioactive substancemeans disintegration of nucleiofthe substance byemissionof different radiations. (b) Despite the strength of the forces that hold nucleons (protons and neutrons) together to forman atomic nucleus,many nuclides are unstable and spontaneouslychange into other nuclides byradioactive decay. (c) The energyliberatedduring radioactive decaycomes fromwithinindividualnucleiwithout externalexcitation, unlike the case of atomic radiation. (d) It is statisticalprocess that obeys the laws of chance. (e) The decayof nucleus takes place to achieve the stable end products. Kinds of Decay There are five kinds of radioactive decays. When radioactivitywas discovered, onlythree kinds of radioactive decays alpha(.), beta(.) and gamma)(.) were known.Whichwere eventuallyidentifiedas 42He nucleus, electronand highenergyphotonrespectively. Later two more kinds of radioactive decays namely positron emission and electron capturewere added. Alpha decay (a) In this type of decay, the unstable nucleus emits an alpha particle. (b) It reduces the proton number Z of the nucleus by 2. (c) It reduces the neutron number Nof the nucleus by 2.

NUCLEAR PHYSICS www.physicsashok.in 8 (d) It reduces themass number (i.e. Z +N) of the nucleus by 4. (e) It changes the element itself hence the chemical symbol of the residualnucleus is different fromthat of the originalnucleus. (f) The alpha decay processmay be represented as A A 4 4 Z Z 2 2 X� . Y He . . . (g) The nucleus before the decay is called the parent nucleus and that obtained after the decay is called the daughter nucleus. Ex: Let us consider the example given below: 212 208 4 83 81 2 Bi�. Tl. He In the above example of .-decay, the parent nucleus is bismuth (Bi) and the daughter nucleus is thallium (Tl). (h) Alpha decay occurs in all nucleiwithmassA> 210. (i) In this decay, the nucleus decreases itsmass number tomove towards stability. (j) On emission of .-particle, the binding energyper nucleon increases and the residualnucleus tends towards stability. Q-Value for .-decay: If .-decay process is given by AZX A 4 Z 2 . Y . + 42He , then Q-value = . A . . A 4 . . 4 . Z Z 2 2 m X m . Y m He . .. . . .. c2 NOTE: The quantity m. . AZX represent atomic mass of the particle X. Example 55. Aradon nucleus Rn86222, ofmass 3.6 × 10�25 kg, undergoes .-decay. The .-particle hasmass 6.7 × 10�27 kg and energy 8.8 × 10�31 J. (a)What is the resulting nucleus? (b) Find the velocity of recoilof the nucleus. Sol. (a) The atomic number will be reduced by 2 and the mass number by 4. . A = 222 � 4 = 218 and Z = 86 � 2 = 84 The resulting nucleus is 84Po218. (b) mass of resulting nucleus = m1 = 3.6 × 10�25 � 6.7 × 10�27 = 3.533 × 10�25 kg Let v1 = its velocity of recoil Mass of .-particle = m2 = 6.7 ×10�27 kg Velocity of .-particle = v2 = 1/2 2 2E m . . .. .. with E = 8.8 × 10�13 J Now m1v1 =m2v2 . 1/ 2 27 13 1/ 2

2 1 25 1 (2m E) (2 6.7 10 8.8 10 ) v m 3.533 10 . . . . . . . . . . , v1 = 3.1 × 10+5 ms�1

NUCLEAR PHYSICS www.physicsashok.in 9 Beta Decay (a) Beta decay is a process inwhich a neutron is converted into a proton. n p + e� + . (b) It increases the atomic number (Z) of nucleus by 1. (c) It does not alter themass number (A). (d) If a nucleus is formedwithmore number of neutrons than needed for stability, a neutronwill convert itself into a proton tomove towards stability. (e) When a neutron is converted into a proton, an electron and a newparticle named antineutrino are created and emitted fromthe nucleus. (f) The antineutrino is denoted bythe symbol .. It is supposed to have zero restmass like photon. It is chargeless and has quantumnumber ±½. (g) The electron emitted fromthe nucleus is called a beta particle and is denoted bythe symbol .� or �1e0. (h) Astreamof beta particles coming frombulk of unstable nuclei is called beta ray. (i) It is also called betaminus decayas negatively charged beta particles are emitted. (j) The beta decay processmay be represented by AZA A Z 1Y . + e� + . (. n p + e� + . ) AZX A Z 1Y . + .� + . (k) An example of beta decay is 6C14 7N14 + e� + . (antineutrino) Q-value for .-decay: (a) .�decay: If .� -decay process is given by AZX A Z 1Y . + .� + . , then Initial rest mass energy, REi = . . AZe ..m X . Zm .. c2 Final restmass energy, REf= . A . Z 1 e e m Y Zm (Z 1) m . .. . . . .. c2 . Q = REi � REf = . A . . A . Z Z 1 m X m Y . .. . .. c2 Because ofthe largemass, the residual nucleus A Z 1Y . does not share appreciable kinetic energy. Thus, the energyQis shared by the antineutrino and the beta particle. Depending on the fraction taken away by the antineutrino, the kinetic energyof the beta particle can be anything between zero and amaximumvalueQ. Positron emission (.+ decay) (a) .+ decay is a process in which a proton is converted into a neutron with emission of positron (e+) and neutrino (v). p n + e+ + v (b) It reduces the atomic number (Z) of nucleus by 1. (c) It does not alter themass number (A). (d) An isolated proton does not beta decay to a neutron. On the other-hand, an isolated neutron decays to a proton.

NUCLEAR PHYSICS www.physicsashok.in 10 (e) If the unstable nucleus has excess protons than needed for stability, a proton converts itselfinto a neutron. (f) When a proton is converted into a neutron, a positron and a neutrino are created and emitted fromthe nucleus. (g) The neutrino is denoted by the symbol v. It is charge-less particle. (h) The positron (e+) has a positive electric charge equalinmagnitude to the charge on an electrons and has a mass equal to themass of an electron. (i) Positron is called the antiparticle of electron. (j) When an electron and a positron collide, both the particles are destroyed and energyismade available. (k) Neutrino and antineutrino are antiparticles of each other. (l) The .+ decay process is represented as AZX A Z 1Y . + e+ + v [. p . n + e+ + v] AZX A Z 1Y . + .+ + v If the unstable nucleus has excess protons than needed for stability, a proton converts itselfinto a neutron. In the process, a positron and a neutrino are created and emitted fromthe nucleus, p n + e+ + v .....(iv) The positron e+ has a positive electric charge equalinmagnitude to the chargeon an electron and has amass equal to the mass of an electron. Positron is called the antiparticle of electron.When an electron and a positron collide, both the particles are destroyed and energy is made available. Similarly, neutrino and antineutrino are antiparticles of each other.When a proton in a nucleus converts itself into a neutron, the decay process is represented as AZX A Z 1Y . + e+ + v or AZX A Z 1Y . + .+ + v .....(v) This process is called beta plus decay. The positron so emitted is called a beta plus particle. . + - decay or position - emission : If the .+-decay or position-emission is given by AZX A Z 1Y . + .+ + v, then R.Ei = . . AZe ..m X . Zm .. c2 R.Ef = . A . Z 1 e e m Y (Z 1)m m . .. . . . .. c2 . Q = R.Ei � R.Ef = . A . . A . Z Z 1 e m X m Y 2m . .. . . .. c2 Q = . A . . A . Z Z 1 e m X m Y 2m . .. . . .. c2 .....(vi) Can an isolated proton decayto a neutron emitting a positron and a neutrino as suggested byequation (iv)?

Themass ofa neutron is larger than themass ofa proton and hence theQ-value of such a processwould be negative. So, an isolated proton does not beta decay to a neutron.On the other hand, an isolated neutron decays to a proton as suggested by equation (i). (m) The positronis also called beta plus particle. (n) An example of .+ decay is 29Cu64 28Ni64 + e+ + v

NUCLEAR PHYSICS www.physicsashok.in 11 Asimilar process, known as electron capture, takes place in certain nuclides. In this process, the nucleus captures one of the atomic electrons (most like an electron fromthe K shell). Aproton in the nucleus combineswith this electron and coverts itselfinto a neutron.Aneutrino is created in the process and emitted fromthe nucleus. Electron capture (a) When the nucleus has toomanyprotons relative to the number ofneutrons, the nucleus captures one of the atomic electrons (most likely an electron fromtheK-shell). (b) Aproton inthe nucleus combineswith this electron and converts itselfinto a neutron. (c) Aneutrino is created in this process and emitted fromthe nucleus. p + e� n + v (d) In this process, atomic number (Z) of the nucleus decreases by 1. (e) This process does not alter the value ofmass number (A). (f) When an atomic electron is captured, a vacancy is created in the atomic shell and X-rays are emitted following the capture. (g) This process is also calledK-capture. (h) The processmay be represented as AZX + e� A Z 1Y . + v [. p + e� n + v] (i) An example of electron capture is 29Cu64 + e� 28Ni64 + neutrino Q-Value of K-capture process: IfK-capture process is given by AZX + e� A Z 1Y . + v, then Q = . A . . A . Z Z 1 m X m Y . .. . .. c2 Example 56: Calculate theQ-value in the following decays- (a) 19O 19F + e + . [.� -decay] (b) 25Al 25Mg + e+ + v, [.+-decay] The atomic masses of 19O = 19.003576 a.m.u. 19F = 18.998403 a.m.u, 25Al = 24.990432 a.m.u.,25Mg = 24.985939 a.m.u. Sol: (a) The Q-value of .� -decay is Q = [m (19O) � m(19F)]c2 = [19.003576 a.m.u. � 18.998403 a.m.u.] (931.5MeV/a.m.u.) = 4.819MeV (b) The Q-value of .+ -decay is Q = [(mass of 25Al nucleus) � (mass of 25Mg nucleus) � (mass of positron)]c2 = [(24.990432 a.m.u � 13me) � (24.985939 a.m.u. � 12 me) � me]c2 = [(0.004593 a.m.u.) � 2me]c2 = 0.004593 a.m.u. × (931.5MeV/a.m.u.) � 2 × 0.511 2 MeV c .c2 = 4.276MeV � 1.022 MeV = 3.254MeV

NUCLEAR PHYSICS www.physicsashok.in 12 Gamma Decay (a) Nucleus has also energylevels like atoms have. (b) This decayprocess is related to the transitions between two nuclear energy levels. (c) The protons and neutrons inside a nucleusmove in discrete quantumstateswith definite energies. (d) In the ground state, the nucleons occupy those quantumstates whichminimise the total energy of the nucleus. (e) The higher energystates are also available to the nucleons and if appropriate energyis supplied, the nucleus maybe excited to higher energies. (f) The energydifferences in the allowed energylevels of a nucleus are generally large (in the order ofMeV). (g) It is difficult to excite the nucleus to higher energylevels byusualmethods of supplying energylike heating etc. (h) When an alpha or a beta decay takes place, the daughter nucleus is generallyformed in one of its excited states. The daughter nucleus in an excited state eventuallycomes to its ground state byemitting one photon ormore than one photon of electromagnetic radiation. (i) The process of a nucleus coming down to a lower energylevelbyemitting a photon is called gamma decay. (j) In this decay, atomic number (Z) aswell asmass number (A) of the nucleus remain constant. (k) In this decay, the quantumstates of the nucleons vary. (l) The electromagnetic radiationemitted in nuclear transitions is called gamma ray. (m) Thewavelength ofthis radiation is given bythe common relation. . = hc E where E = energy of the photon. (n) An example of gamma decayis shown in figure below: . . . 57Co .+ 136 keV 14 keV 0 keV 2nd excited state 1st excited state Ground state 57Fe When 57Co is taken in bulk, we can observe a streamof .+ particles, 136 keV photons, 122 keV photons and 14 keVphotons coming fromthe 57Co source. NOTE: The ., . and . rays are collectively called nuclear radiation. Comparison among the kinds of decay (a) The velocityof .-particles is relatively low: v. = (c/30 � c/15),when c is the velocity of light Mass m. = 4 amu charge q. = +2e For ..-particles: 0 . v. < c m. =mass of an electron q. = �e

NUCLEAR PHYSICS www.physicsashok.in 13 For .-rays: Since .-rays are electromagneticwaves, hence theypropagatewith the speed of light. i.e. v. = c Rest mass,m. = 0(like photon) q. = 0 (like photon) (b) The penetrability of .-rays is 0 - 100 times higher than the penetrability of b-rays and 1000 - 10000 times higher than the penetrability of .-rays. The penetrability of .-rays also exceeds the penetrabilityof x-rays. Carboard Al Lead .. . . . - particles from radioactive materials are stopped by a piece of cardboard. -particles penetrate the cardboard but are stopped by a sheet of aluminium. Even a thick slab of lead may not stop all the -rays. (c) In a magnetic field, a beamof ., . and .-rays splits into three parts. In amagnetic field, .-rays are undeviated and .-particles aremost deviated .-particles .-particles .-rays Magnetic field (d) Table for various decay directed into paper Decay Transformation Example .-decay AZX A 4 4 Z 2 2 . Y He . . 238 92U 234 4 90 2 Th . He .-decay AZX A Z 1Y . + e� + . 146C 147N + e� + . Positron emission AZX A Z 1Y . + e+ + v 64 29Cu 64 28Ni + e+ + v Electroncapture AZX + e� A Z 1Y . + v 64 29Cu + e� 64 28Ni + v Gamma decay A * Z X AZX + . 87 * 38Sr 87 38Sr + . The *denotes anexcited nuclear estate, .-denotes a gamma-rayphoton and v and . denotes neutrino and antineutrino particles respectively. C52:An a-particle is bombarded on 14N.As a result, a 17O nucleus is formed and a particle is emitted. This particle is a/an

(a) neutron (b) proton (c) electron (d) positron

NUCLEAR PHYSICS www.physicsashok.in 14 Sol: 14 4 7 N . 2He 17 1 8 1 O . p 11p is equivalent to 11H . C53: In a radioactive decay, neither the atomic number nor themass number changes.Which of the following particles is emitted in the decay ? (a) Proton (b) Neutron (c) .- particle (d) photon Sol: Photon is equivalent to .-radiation. Inthis decay, onlyquantumstates of the nucleons vary. C54: .-rays emitted bya radioactivematerial are (a) electromagneticwaves (b) electrons orbiting around the nucleus (c) charged particles emitted by the nucleus (d) neutral particles Sol: Aneutronin the nucleus decays emitting an electron. C55: Give an equation representing the decay ofa free neutron. Sol: 10 n 1 0 1 1 H e v. . . . C56: Howmanyelectrons, protons and neutrons are there in a nucleus of atomic number 11 andmass number 24 ? Sol: Number of electrons or protons, Z= 11 and number of neutrons, N=mass number �Atomic number = 24 � 11 = 12 C57: Fill up the blanks (i) hv e� + .............. (.-photon) (electron) (ii) 90Th234 �1B0 + 91Pa234 + ............. (Thorium) (.-particle) (Protactinium) (iii) 92U238 2He4 + ................ (Uranium) (.-particle) Sol: (i) e+(Positron) (ii) . (Antineutrino) (iii) 90Th234(Thorium) C58: Following the origin of gamma decay, calculate the value of wavelength of resulting photonwhen the nucleus of 27 13Al reaches the ground state (Eg = 0) from the state in which Eex. = 1.015MeV.The related figure is shown adjacent. . .� 1.015 MeV 27Mg 12 0 MeV 27Al Sol: .(in Å) = 13 12400 .E(eV) = 6 12400 (1.015 . 0).10 = 0.012217 Å LAW OF RADIOACTIVE DECAY When the radioactive substance is only disintegrating: Radioactive decay is a randomprocess. Each decay is an independent event, and on

e cannot tellwhen a particular nucleuswilldecay.When a given nucleus decays, it is transformed into another nuclidewhichmay

NUCLEAR PHYSICS www.physicsashok.in 15 or may not be radioactive. When there is a very large number of nuclei in a sample, the rate of decay i.e. dN dt . . . . . . . is proportional to the number of nuclei, N, that are present i.e. � dN dt . N . � dN dt = .N where . is called the decay constant. This equationmay be expressed in the form dN dt = �.dt and integrated to get 0 N N dN N . = �. t0 . dt , to yield, ln 0 N N = �.t, where N0 is the initial number of parent nuclei at t = 0. The number that survive at time t is therefore N = N0e�.t Definition of decay constant : The probability of decay per second for a particular process for a sample is called the decay constant for that process for that sample. Radioactivity lawof decay gives, . = dN Ndt . Half life : The time period for the number of parent nuclei to fall to 50%is called the half-life, T, andmay be related to .. ifwe put N =N0/2 at t = T, the exponential decay equation gives, 0.5N0 = N0e�.T . .T = ln|2| = 0.693 , T = 0.693 . (a)It takes one half-life to drop to 50%of anystarting value. (b) The half-life for the decayof the free neutron is 12.8min. (c) Other half lives range fromabout 10�20s to 1016 years. Mean life(Tm) : Mean life of a radioactive sample is defined as the average of the lives of all nuclei. . Tm = 0 0 N 0 tdN

N . = 1. = T 0.693 . Tm = 1. and T = 0.693 Tm NOTE : (i) Radioactive decay equation, N = N0e�.t, can also be written as N = N0 t /T 12 . . . . . . , where T is half-life of the sample. (ii) The decay constant is also given as, . = 2.303 t log 0 NN

NUCLEAR PHYSICS www.physicsashok.in 16 Activity of radioactive substance: Since the number of atoms is not directlymeasurable, wemeasure the decay rate or activity (A) A= � dN dt , but N = N0e�.t A = � dN dt = .(N0e�.t) = .N . A = .N =A0e�.t, whereA0 = .N0 is the initial activity. BothNandAdecrease exponentiallywith time. The activityis characterized by the same half-life. Units of activity: (a) The SI unit for the activityis the becquerel (Bq), but the curie (Ci) is often used in practice. 1 becquerel (Bq) = 1 disintegrations per second (dps) 1 curie (Ci) = 3.7 × 1010 dps 1 rutherford = 106 dps (b) Rate of decayof 1 gmsubstance is called specific activity. (c) Activity of 1 gmRa226 is 1Ci. Example 57: The half-life of 198Au is 2.7 days. Calculate (a) the decayconstant, (b) the average life and (c) the activity of 1.00mg of 198Au. Take atomicweight of 198Au to be 198 g/mol. Sol: (a) The half-life and the decay constant are related as T = 0.693 . . . = 0.693 T = 0.693 2.7 days = 0.693 2.7.24.3600s . . = 2.9 × 10�6 s�1. (b) Tm= T 0.693 = 2.7 0.693 days = 3.9 days. (c) Activity is given as,A= .N, whereNis the number of nuclei present in 1mg of 198Au. Atomic mass of 198Au = 198 g . N = 1 mg 198g ×Avogadro no . N = 10 3 198. × 6 × 1023 atoms . N = 3.03 × 1018 atoms Thus, A = .N = (2.9 × 10�9 s�1) × 3.03 × 1018 (atoms) = 8.8 × 1012 disintegrations per sec. = 12 10

8.8 10 3.7 10 .. Ci = 240 Ci

NUCLEAR PHYSICS www.physicsashok.in 17 Example 58. Two radioactive materialA1 andA2 have decayconstants of 10 .0 and .0. If initially they have same number ofnuclei, the ratio of number of their undecayed nucleiwill be (1/e) after a time (A) 0 1 . (B) 0 1 9. (C) 0 1 10. (D) 1 Sol. 1t 10 0t 1 0 0 N . N e.. . N e. .2t 0t 2 0 0 N . N e.. . N e.. 0 0 10 t 1 0 t 2 0 N N e N N e. . .. . 1 9 0t 2 N e N . . . . 9 0t 1 e e . . . e.1 . e.9.0t �1 = �9.0t 0 t 1 9 . . Hence (B) is correct. C59. Acertain element has a density of 10 g cm�3 and half-life of 140 days. Over a period of 140 days, the average number of .-emissions per day is found to be 12 × 1012, from a sample of initial mass 1 µg. Estimate the number of atoms in1 cm3 ofthe element. Sol. We assume that only one emission takes place per atom. In 140 days, no. of emissions = 140 × 12 × 1012 . initial no. of atoms present = 2 × 140 × 12 × 1012 (since 140 days is the half-life) . no. of atoms in 1 µg = 28 × 12 × 1013 . no. of atoms in 1 cm3, i.e., 10g = 3.36 × 105 × 107 = 3.36 × 1022 C60:Aradioactive sample has 3.2 × 1016 active nuclei at certain instant.Howmany of

these nucleiwill stillbe in the same active state after four half-lives ? Sol: In one half-life the number of active nuclei reduces to half the original number. Thus in four half-lives the number is reduced to 12 . . . . . . 12 . . . . . . 12 . . . . . . 12 . . . . . . i.e. 1 16 th of the original number. . The number ofremaining active nuclei is, = 1 16 × 3.2 × 1016 = 2 × 1015 nuclei

NUCLEAR PHYSICS www.physicsashok.in 18 C61: The activityof a radioactive sample falls from1200 s�1 to 1000 s�1 in 60minutes. Calculate its half life. Sol:We have, A=A0e�.t (1000 s�1) = (1200 s�1) e�.t . 5/6 = e�.t .t = ln(6/5) . . = ln(6 / 5) t = ln(6 / 5) 60min but, T = ln 2 . . ln 2 T = ln(6 / 5) 60min T = ln 2 ln(6 / 5) × 60 min = 0.693 0.182 × 60 min . half-life = 228min C62.At a given instant there are 25%undecayed radioactive nuclei in a sample.After 10 second themumber of undecayed nuclei reduces to 12.5%Calculate : (i)mean life of the nuclei (ii) the time inwhich the number of undecayed nucleiwill further reduce to 6.25%ofthe reduced number. Sol. (i) In 10 second, number of nuclei has been reduced to half (25%to 12.5%). Therefore, its half life is t1/2 = 10 s Relation betweenhalf life andmean life is 1/ 2 mean t t 10 s ln (2) 0.693 . . tmean = 14.43 s (ii) Frominitial100%to reductiontill6.25%, it takes four half lives. t1/ 2 t1/ 2 t1/ 2 t1/ 2 100%...50%...25%...12.5%...6.25% . t = 4 t1/2 = 4(10) s = 40 s Example 59.Asample initiallycontains 1020 radioactive atoms of half-life 130days. Calculate the activityof the sample after 260 days have elapsed.Also find the total energy released during this period if the energy released per disintegration is 8 × 10�13 J. In 260 days, i.e., two half-lives, the number of undisintegrated atomswill reduce to 1/4th. Sol. . no. of atoms present after 260 days = N = 14 × 1020 Also, disintegration constant = . = 0.6931/T where T = 130 days = 130 × 86400 s Now, activity = .N = 0.693 130 . 86400 ×

14 × 1020 = 1.54 × 1012 s Number of atoms present initially= 1020 Number of atoms present after 260 days = 14 × 1020 . number of disintegrated atoms = 3/4 × 1020 Energy per disintegration = 8 × 10�13 J . total energy released = 34 × 1020 × 8 × 10�13 = 6 × 107 J

NUCLEAR PHYSICS www.physicsashok.in 19 Example 60. Abodyofmassm0 is placed on a smooth horizontal surface. Themass of the body is decreasing exponentiallywith disintegration constant ..Assuming that themass is ejected backwardswith a relative velocity u. Initially the bodywas at rest. Find the velocityof it after time t. Sol. Mass of the bodyleft after time t is m = m0e�.t So dm dt . . . .. .. = m0.e�.t and thrust force on the body is Ft = ur dm dt . . . .. .. (in forward direction) or m dv dt . . .. .. = u(m0.e�.t) (ur = u) or (m0e�.t) dv dt = m0 u.e�.t or dv = u.dt or v t 0 0 . dv . u.. dt or v = u.t Example 61. Aradio nuclidewith half life T = 69.31 second emits .-particles of average kinetic energyE = 11.25 eV.At an instant concentration of .-particles at distance, r = 2mfromnuclide is n= 3 × 1013 perm3. (i) Calculate number ofnuclie in the nuclide at that instant. (ii) If a small circular plate is placed at distance r fromnuclide suchthat .-particles strike the plate normallyand come to rest, calculate pressure experienced bythe plate due to collision of .-particles. [Mass of .-particle = 9 × 10�31 kg)] Sol. Let activity (rate of decay) of the nuclide beAnuclie per second. It means A..-particles are emitted per second. If a spherical surface of radius rwith centre at position of nuclide be considered thenA .-particles cross this surface (v dt) r per second. Itmeans during an elemental time intervaldt a number (A. dt) of .-particles cross this surface. If velocity of .-particles be v then above calculated (A. dt) .-particles are in a space having shape of a spherical shell of radius r and radial thickness (v dt) as shown in figure. Volume of this space = 4.r2 (v dt) . Concentration of .-particles at distance r fromnuclide is 2 n A dt 4 r (v dt) . . or activityofthe nuclide, A = 4.r2 vn But activity, A= .NwhereNis number of nuclei Hence, N =

4.r2vn . but decay constant ..= log 2 T . 4 r2vnT N 0.6931 . . ...(1)

NUCLEAR PHYSICS www.physicsashok.in 20 Kinetic energy of .-particles, E 1 mv2 2 . , v 2E m . substituting this value in equation(1), 2 N 4 r nT 2E 9.6 1022 0.6931 m . . . . . (ii) At distance r fromthe nuclide (A/4.r2) .-particles cross unit area per second. Let area of small circular plate be S, then number of .-particles striking the plate per second 2 2 2 A S N S 0.6931 NS 4 r 4 r 4 r T . . . . . . . Momentumof eachparticle just before collisionismv andafter collisionparticles come to rest ormomentum becomes zero. . Momentumtransferred to plate due to collision is .p = mv � 0 = mv Due to transfer ofmomentum, the plate experiences a forcewhichis equal to rate oftransfer ofmomentum. . Force, F = .p × no. of particles striking per second or F = mv × 2 0.6931NS 4.r T Pressure, P = Force per unit area . P = FS = 2 0.6931N 4.r T mv but v = 2E m . P = 2 0.6931N 2mE 4.r T = 1.08 × 10�4 Nm�2 RADIOACTIVE DATING OR CARBON DATING Radiocarbon dating, also called carbon dating, is used to estimate the age oforganic samples. The technique is based on the .-activity of the radioactive isotope C14; 14 6 C 147N + .� + . ..� . beta particles, . . anti neutrino] (a) High energyparticles for outer space, called cosmic rays, induce nuclear reactions in the upper atmosphere and create carbon - 14. (b) The carbon dioxidemolecules of the earth�s atmosphere have a constant ratio (~.1.4 × 10�12) ofC14 to C12

isotopes. (c) All living organisms also showthe same ratio as theycontinuously exchange CO2withtheir surroundings. (d) However after its death, an organismcan no longer absorbCO2 and the ratio C14/C12 decreases due to the .-decay of C14. (e) Thus bymeasuring the .-activityper unit mass, it is possible to estimate the age of amaterial. (f) Half life of 14C is 5739 y.

NUCLEAR PHYSICS www.physicsashok.in 21 Example 62:When charcoalis prepared froma living tree it shows a disintegration rate of 15.3 disintegrations of 14C per gram per minute. Asample from an ancient piece of charcoal shows 14C activity to be 12.3 disintegrations per gramperminute. Howold is this sample ?Half life of 14C is 5730 y. Sol: 14C-activity of a living tree,A0 = 15.3 dis. per min. 14C-activity of the old sample, A= 12.3 dis. permin. Suppose, the sample is t year old, then (14C-activity of sample) = (14C-activity of living tree )e�.t i.e. A=A0e�.t .....(i) where . is decayconstant of 14C-activity Putting values ofAandA0 in equation (i),we get 12.3 = 15.3 e�.t ....t = ln 15.3 12.3 . ...t = ln(1.24) = 0.218 . t = 0.218 . = 0.218 0.693 T, [as, . = 0.693 T ] . t = 0.218 0.693 × 5730 y, [T = half-life of 14C -activity= 5730 y] . t = 1805 y Thus, the sample is 1805 y old. Example 63: The relative radiocarbon activityin a piece of charcoal fromthe remains of an ancient camp fire is 0.18 that of a contemporary specimen.Howlong ago did the fire occur ?Half-life of 14C-activity is 5730 y. Sol: Here, ratio between the 14C-activityof burnt charcoal and that of a living tree is given. 14 14 0 C activity of charcoal (say,A) C activity of a living tree (say,A ) . . = 0.18 . 0 A A = 0.18 Suppose, t year ago fire occurred and let . be the decayconstant of 14C-activity.We have, using radioactive decay law, A=A0e�.t ...e.t = 0 AA . .t = ln 0 AA . t = 1. ln 0 AA . t = T 0.693 ln 0 AA , [. =

0.693 T ] = 5730 0.693 ln 1 0.18 = 8268.3 ln|5.56| = 1.4 × 104 y Thus, fire occurred 1.4 × 104 years ago. Radioactivity law for different types of disintegration of the radioactive substance It is seenthat radioactive disintegration of a radioactive substance is not only ruled bythe radioactive law, N=N0e�.t (whichis applicable onlywhenthe radioactive substance is onlydisintegrating), but the radioactive lawchanges for various types ofdisintegrationofthe substance.Let us first enlist these types ofdisintegration as:

NUCLEAR PHYSICS www.physicsashok.in 22 (1) Radioactive substance only disintegrates: For this radioactive law is � dN dt = .N or N = N0e�.t (2) Disintegration with continuous production of the radioactive substance. It deals with the case when production and the decay of the radioactive substance are taking place simultaneously. Formation A decays (.) B (3) Successive disintegrations of the products: It dealswith the casewhen a substanceAdecays into a substance Band Bsuccessively decays into a third substance Cwith the same or different decay rates. B decays (. ) 2 decays C (. ) 1 A (4) Simultaneous disintegrations of parent nuclei: It dealswith the casewhen a parent nucleusmaydisintegrate in a number ofways into different products. A . B 1.2 C Parent nucleusAmay decay in B or Cwith decay constants .1 and .2 respectively. (5) Radioactive equilibrium: In a radioactive series, after a period of time, successive daughter nuclei decay at the same rate as it is formed. This situationis called radioactive equilibrium. (6) Disintegration of isotopes: Apreparationmayhave a number ofradioactive isotopes.Herewewilldealwiththe net rate ofdisintegration of the preparation. Let us discuss these types of disintegrations. Radioactive substance only disintegrates : Suppose disintegration ofAinto Bis taking placewithdecay constant ., then decayrate, � dN dt = .N . N = N0e�.t Disintegration with continuous production: Suppose a substanceAdecays into B with decay constant . and simultaneously the production ofAis taking place at a constant rate q. decays B (.) Formation (q) A Let, Nis the number of nuclei ofApresent at time t. q = constant rate of formation ofA. Disintegration ratewillbe given by � dN dt = �q + .N, where .N is the rate of decay ofA.

Rearranging,we get,

NUCLEAR PHYSICS www.physicsashok.in 23 dN .q . .N = �dt, Integration gives, 0 N N dN . .q . .N = � t0 .dt ,whereN0 is the no. of nuclei ofAinitially present. Finally, N = 1. [q + (.N0 � q)e�.t] Example 64. Aradionuclidewith disintegration constant . is produced in a reactor at a constant rate . nuclei per second.DuringeachdecayenergyE0 isreleased. 20%ofthis energyisutilised inincreasingthe temperature ofwater. find the increase in temperature ofmmass ofwater in time t. Specific heat ofwater is s.Assume that there is no loss of energythroughwater surface. Sol. Let Nbe the number of nuclei at any time t. Then Rate of formation = N Rate of decay = N . net rate of formationof nuclei at time t is dN dt = . � .N or N t 0 0 dN dt N . . . . . . or N = .. (1 � e�.t) Number of nuclei formed in time t = .t and number of nuclei left after time t .1 e..t . . . . . Therefore, number of nucleidisintegrated in time t t .1 e..t . . . . . . . . energy released till time . t . 0 t E t 1 e.. . . . . .. . . . . . . . But only20%of it is used in raising the temperature ofwater. So . t . 0 0.2 E t 1 e.. Q . . . .. . . . . . . . where Q = ms.. . ...= increase in temperature ofwater = Q ms . . t . 0 0.2 E t 1 e ms .. . . . .. . .. . .. .. . Example 65. Nuclei of a radioactive element Aare being produced at a constant ra

te .. The element has a decay constant ..At time t = 0 , there areN0 nucleiof the element. (a) Calculate the number Nof nuclei ofAat time t (b) If . =2N0., calculate the number of nucleiofAafter one half life ofA, and also the limiting value ofN as t ....

NUCLEAR PHYSICS www.physicsashok.in 24 Sol. (a) Let at time �t�, number of radioactive nuclei areN. Net rate of formation of nucleiofA. dN dt = . � .N or dN . . .N = dt or 0 N t N 0 dN dt N . . . . . Rate of formation = A Rate of decay = N t = t N = N Solving this equation,we get . . t 0 N . 1 .. . . . .N e.. . . . . ...(1) (b) (i) Substituting . = 2.N0 and t = t1/2 = ln (2) . in equation (1), we get 0 N 3 N 2 . (ii)Substituting . = 2.N0 and t ... in equation (1), we get N = .. = 2N0 or N = 2N0 Example 66. Aradionuclide with half life T is produced in a reactor at a constant rate q nuclei per second. During each decay, energyE0 is released. If production of radionuclide is started at t = 0, calculate. (i) rate of release of energyas function of time t and (ii) total energy released upto time t. Sol. To calculate rate of release of energy at time t and total energy released upto time t, rate of decayat that instant and totalnumber of decays upto that instantmust be known. Since, nucleiproduced are radioactive, therefore, their decay starts as soon as their production is started. Let at some instant number of nuclei in the radionuclide beN. Then rate ofits decay= .Nwhere . is decay constant which is equal to e log 2 T . . .. .. . Since, rate of production is q nuclei per second, therefore, at instant t, net rate of increase of nuclei e dN q N q Nlog 2 dt T . . . . . or e dN qT Nlog 2 dt T .

.

. e dN dt qT N log 2 T . . ...(1) Integrating above equationwith limits at t = 0, N= 0 and at t,N = ? N t 0 0 e dN dt qT N log 2 T . . . . . t loge 2 T e qT N 1 e log 2 . . . . . . . . .

NUCLEAR PHYSICS www.physicsashok.in 25 Hence, rate of decay, t loge 2 A N q 1 e T . . . . . . . . . . . Since, energyE0 releases during each decay, therefore, rate of release of energy at time t loge 2 T 0 0 t AE qE 1 e. . . . . . . . . . Total number of nuclei produced upto time t = q . t But the number of nuclei remaining undecayed at that instant isN. Therefore, total number of nucleiwhich decayed upto time t = (qt � N) Hence, total energy released upto this time = (qt �N)E0 t loge 2 0 T 0 e qTE qtE 1 e log 2 . . . . . . . . . . Successive disintegration : Suppose a radioactive substanceAdecays into B with decay constant .1 andB successivelydecays into another stable product Cwith a decay constant .2. decays C (stable product) .2 B .1 A decays Let N0 be the number of nuclei ofApresent at t = 0, N1, N2, N3 be the number of nuclei ofA, B and C respectively at anyinstant t. Decay rate ofAis given by dN dt . = .1N1 . N1 = N0e�.1t .....(i) Rate of change of no of nucleiofB is 2 dNdt = (Rate of decay ofA) � (Rate of decay of B) But, Rate of decay ofA= .1N1 and rate of decay of B = .2N2 hence, 2 dNdt = .1N1 � .2N2 .....(ii) Rate of change of number of nuclei ofC is 3 dNdt = (rate of decay ofB) = .2N2 . 3 dNdt = .2N2 .....(iii) Solving (i), (ii) and (iii),we get N1 = N0e�.1t N2 = 0 1 2 1 N .

. . . [e�.1t� e�.2t] N3 = N0 2t 1t 1 2 2 1 e e 1 .. .. . . .. . . . . . . . . .

NUCLEAR PHYSICS www.physicsashok.in 26 NOTE : In this case total number of nuclei remains constant, hence, N1 + N2 + N3 = N0 at any time. Example 67:Aradioactive nucleusAdecays to a nucleus B with a decay constant .1. B further decays to a stable nucleusCwith a decayconstant .2. Initiallythere are onlyAnuclei and their number isN0. Set up the rate equations for the populations ofA,BandC.The population ofBnucleus as function of time is given by N2(t) = 0 1 1t 2t 2 1 N (e e ) ( ) .. .. . . . . . . . . . . . . Calculate the population ofCas a function of time t. Sol: Let N1, N2 and N3 be the number ofA, B and C nuclei respectively present at a time t. Decay rate forAnucleiwill be � 1 dNdt = .1N1 . N1 = N0e�.1t Rate of change of the number of nucleiofB is 2 dNdt = .1N1 � .2N2 where .2N2 = decay rate of B and .1N1 = decay rate ofA Rate of change of the number of nuclei ofC is 3 dNdt = (Rate of decay of B) = .2N2 3 dNdt = .2N2 . N3 3 0 . dN = t 2 2 0 . . N dt , after integrating. . N3 = .2 t 2 0 . N dt but, N2 is given as, N2 = 0 1 1t 2t 2 1 N (e e ) ( ) .. .. . . . . . . . . . . . . N3 = 1 2 t 0 1 2 t t 2 1 0 N . . (e.. e.. )dt . . . . . = 0 1 2 2 1 N . . . . . 1 2

t t t 1 2 0 . e.. e.. . . . . . .. .. . = 0 1 2 2 1 N . . . . . 1t 2t 1 2 1 2 . e.. e.. . 1 1 .. . . . . .. . .. .. . .. . .. . N3 = 0 2 1 N . . . 2 t 1t 1 2 [. e.. . . e.. ] + N0 . N3 = N0 2t 1t 1 2 2 1 e e 1 .. .. . . .. . . . . . . . . . Simultaneous disintegration: Aradioactive nucleus can decay bytwo different processes. For example a nucleusAmay either a-decay to a nucleusB or .-decay to nucleus C. A . B 1.2 C Let .1 and .2 be the decay constants for these two decay processes.

NUCLEAR PHYSICS www.physicsashok.in 27 The probability that an active nucleus decays by the first process in a time interval dt is .1dt .As decay constant, . is defined as the probabilityofdecayper second for a particular process for a sample. Similarly, the probability that it decays by the second process is .2dt. Hence, the probability the it either decays by the first process or by the second process is .1dt + .2dt. If the effective decay constant is .eff, this probability is also equal to .effdt. Thus, .effdt = .1dt + .2dt . .eff = .1 + .2 For a number of different process for decay, .eff = .1 + .2 + ............ Example 68:Aradioactive nucleus can decay bytwo different processes. The half life for the first process is t1 and that for the second process is t2. Showthat the effective half-life t of the nucleus is given 1t = 1 1t + 2 1t . Sol: The decay constant for the first process is .1 = 1 ln 2 t For the second process, .2 = 2 ln 2 t Probability of decay by the first process = .1dt Probability of decay by the second process = .2dt Probability that it either decay by the first process or the second process = .1dt + .2dt This probability also equals to .effdt,where .eff is the effective decay constant. Thus, .effdt = .1dt + .2dt . .eff = .1 + .2 . ln 2 t = 1 ln 2 t + 2 ln 2 t . 1t = 1 1

t + 2 1t Proved. Example 69. Anumber N0 of atoms of a radio active element are placed inside a closed volume.The radiactive decayconstant for the nucleus of this element is .1. The daughter nucleus that formas a result of the decay process are assumed to be radioactive toowitha radioactive decayconstant .2.Determine the time variation of the number of such nucleus. Consider two limiting cases .1>> .2 and .1<< .2. Sol. In time intervaldt, number of increase ofdaughter nuclei are dN2 = .1N1dt � .2N2dt or dN2 = .1N0 e..1t dt � .2N2dt (N1 = N0 e..1t ) or 2 dNdt + .2N2 = .2N0 e..1t ...(1) Case-1 : When .1 > >..2 i.e. (t1/2)1 < < (t1/2)2 (t1/2 = half life) We can assume that N20 . N0 so that N2 = N0 e..2t (N20 = number of daughter atoms at time t = 0)

NUCLEAR PHYSICS www.physicsashok.in 28 Physicallythismeans that parent nucleipractivallyinstantlytransforminto daughter nuclei,which thendecay according to the lawof radioactive decaywith decay constant .2. Case-2 : When .1 < <..2 i.e. (t1/2)1 > > (t1/2)2 In this case number of parent nuclei can be assumed to remain constant over a sizable time interval and is equal to N0. This transforms equation (1) into 2 dNdt = � (.2N2 � .1N0) or N2 t 2 0 1 0 2 2 0 dN dt N N . . . . . . Which after integration gives N2 = 12 .. N0(1 � e..2t ) Example 70. Aradio nuclide consists of two isotopes.One of the isotopes decays by .-emission and the other by .-emissionwith half livesT1 = 405 second and T2 = 1620 second, respectively.At t = 0, probabilities of getting ..and ..particles fromthe radionuclide are equal. Calculate their respective probabilities at t = 1620 second. If at t = 0, total number of nuclie in the ratio-nuclide areN0, calculate time t when total number of nuclie remained undecayed becomes equal to 0 N2 . Given log10 2 = 0.30103, log10 13 = 1.11394 Sol. Since, at t = 0, probabilities of getting . and . particles fromthe radionuclide are equal, therefore, initial activities of two isotopes are equal. Let it beA0. Activity of first isotope at t = 1620 sec. t /T1 0 1 0 1 A A A 2 16 . . . . .. .. That of second isotope, t /T2 0 2 0 1 A A A 2 2 . . . . .. .. . Total activity of radionuclide at t = 1620 sec,A=A1 +A2 = 9 16 A0 . Probabilityofgetting .-particle, 1 1

P A 1 A 9 . . and that of getting .-particle, 2 2 P A 8 A 9 . . Let at t = 0, number of nuclei of two isotopes beN01 andN02, respectively. Initial activityof first isotope, 1 01 1 01 1 A N N log 2 T . . . That of second isotope, 2 02 2 02 2 A N N log 2 T . . .

NUCLEAR PHYSICS www.physicsashok.in 29 Since, A1 =A2, therefore 01 02 1 2 N N T T . or 01 1 02 2 N T 1 N T 4 . . Initially, total number of nuclei, N0 =N01 +N02 . 01 N 15 . N0 and 02 N 45 . N0 At time t, number of nucleiof first isotope that remain undecayed, t /T1 t / 405 1 01 0 N N 1 1 N 1 2 5 2 . . . . . . .. .. .. .. That of second isotope, t /T2 t /1620 2 02 0 N N 1 4 N 1 2 5 2 . . . . . . .. .. .. .. . Total number of nuclei remaining undecayed at time t, N = N1 + N2 = t / 405 0 N 1 5 2 . . .. .. + t /1620 0 4 N 1 5 2 . . .. .. = 4 t /1620 0 N 1 4 1 5 2 2 .. . . . . .. . . . . . ... . .. . . But it is equal to 0 N2 . . 4 t /1620 0 0 N 1 4 1 N 5 2 2 2 .. . . . . .. . . . . . . ... . .. . . or t /1620 1 8 2 13

. . . .. .. Taking log, t 1620 . log 2 = log 8 � log 13 or .log13 log8. 1620 t s log 2 . . . or .log13 3log 2. 1620 t s 1134 s log 2 . . . . Radioactive equilibrium : Decayof 238 92 U into a stable end product 206 82 Pb is aradioactive serieswhichcontains anumber ofintermediate members. The intermediatemembers of each decayseries havemuch shorter half-lives than their parent nuclide.As a result, ifwe start with a sample NA nuclei of a parent nuclideA, after a period of time an equilibriumsituationwill come about inwhich each successive daughter B, C, ........... decays at the same rate as it is formed. Thus the activities RA, RB, RC, ........... are all equal at equilibrium, and sinceR = .Nwe have.

NUCLEAR PHYSICS www.physicsashok.in 30 .ANA = .BNB = .CNC = ........... This situation is called radioactive equilibrium. The above equation can be used to establish the decay constant (or half-life) of anymember ofthe series if the decayconstant of anothermember and their relative proportions in a sample are known. Example 71: The atomic ratio between the uraniumisotope 238Uand 234Uin amineral sample is found to be 1.8 × 104. The half-life of 234U is 2.5 × 105 year. Find the half-life of 238U. Sol: The two isotopes are in radioactive equilibrium.Hence activities of the twowillbe equal. Thus, .1N1 = .2N2, where, .1, .2 are decay constant of 238U and 234Urespectively. N1 and N2 are number of atoms of 238U and 234U respectively, we have .2 = 5 ln 2 2.5.10 (year)�1, 12 NN = 1.8 × 104, .1 = ln 2 T , where T is half life of 238U. Now, .1N1 = .2N2 ... ln 2 T 12 NN . . . . . . = 5 ln 2 2.5.10 (year)�1 . T = 12 NN . . . . . . × 2.5 × 105 year = 1.8 × 104 × 2.5 × 105year . half life of 238U = 4.5 × 105 year Disintegration of isotopes : Suppose a sample is a mixture of three radioactive isotopesA, B and C. Let .1, .2 and .3 be the decay constant ofA, B and C respectively. N1, N2 and N3 be the number of nuclei of isotopesA, B and C respectively at any instant. IfA1,A2,A3 are decay rates ofA, B and C respectively then, the net decay rate is Anet = A1 + A2 + A3 but, A = .N .. .net N = .1N1 + .2N2 + .3N3 . .net = 1 1 2 2 3 3 N N N

N . . . . . Nis the total no. of nuclei in the sample at any instant . N = N1 + N2 + N3 hence net decay constant, .net = 1 1 2 2 3 3 1 2 3 N N N N N N . . . . . . . Example 72:Asample of uraniumis amixture of three isotopes 234 92 U, 235 92U and 238 92U present in atomic ratio of 0.006%, 0.71%and 99.284%respectively.The half life of these isotopes are 2.5 × 105 years, 7.1 × 108 years and 4.5 × 109 years respectively. Calculate the contribution of activity (in%) of each isotope in this sample.

NUCLEAR PHYSICS www.physicsashok.in 31 Sol: Let N1, N2 and N3 be the number of the three isotopes in the sample. we haveN1 : N2 : N3 = 0.006 : 0.71 : 99.284 .....(i) If .1, .2, .3 are decay constants of these isotopes and as activityA= .N, we have A1 :A2 :A3 = .1N1 : .2N2 : .3N3 A1 :A2 :A3 = 1 1 (ln 2)N T : 2 2 (ln 2)N T : 3 3 (ln 2)N T , [as . = ln 2 T ] . A1 :A2 :A3 = 1 1 NT : 2 2 NT = 3 3 NT .....(ii) where, T1, T2 and T3 are half lives of the isotopes. Comparing (i) and (ii) we get A1 :A2 :A3 = 1 0.006 T : 2 0.71 T : 3 99.284 T = 5 0.006 2.5.10 : 8 0.71 7.1.10 : 9 99.284 4.5.10 = 60 2.5 : 1 : 99.284 4.5 = 24 : 1 : 22.06 A1 :A2 :A3 = 24 : 1 : 22.06 = 24 100 24 1 22.06

.

. . : 1 100 24 1 22.06 . . . : 22.06 100 24 1 22.06 . . . = 2400 47.06 %: 100 47.06 %: 2206 47.06 = 51.00%: 2.12%: 46.88% Activityratio = 51%: 2.12%: 46.88% . Contribution of 234 92 U = 51%; Contribution of 235 92 U = 2.12%; and Contribution of 238 92 U = 46.88%in activityof the sample. Example 73. The isotopes of uraniumU238 andU235 occur in nature in the ratio 128 : 1.Assuming that at the time ofthe earth�s formationtheywere in equal ratio,make an estimate ofthe age of the earth.The half-lives of U238 and U235 are 4.5 × 109 years and 7.13 × 108 years, respectively. Sol. Let N0 be the initial number of atoms. Then fromN= N0e�.t, (ln 2/T1)t 1 0 N . N e. and (ln 2/ T2 )t 2 0 N . N e. . ln 2.1/T2 1/T1.t 1 2 N / N e . . ln 2.1/ 7.13 108 1/ 4.5 109 .t 128 e . . . . . 128 eln 2 1.18 10 9 t . . . . or 27 eln 2 1.18 10 9 t . . . . or t 7 1.18 . × 109 = 5.9 × 109 years

NUCLEAR PHYSICS www.physicsashok.in 32 Types of Nuclear Collision: Exoergic collision / reaction: IfQis positive, restmass energyis converted to kineticmass energy(K3,K4 etc), radiation energyor both, and the reactionis called exoergic. Note: (i) The kinetic energyKE. of the emitted .-particle is never quite equal to the disintegration energyQ because the nucleus recoils with a small amount of kinetic energywhen the .-particle emerges (since momentummust be conserved). . * KE A 4 Q A . . . ,Ais themass number of the parent nucleus. (ii) Themass numbers of nearlyallalpha-emitters exceed 210, and hence most ofthe disintegration energy appears as theKE of the .-particle. (iii) In the .�decay process, the energyQis shared bythe antineutrinos and the beta particle. The kinetic energy of the beta particle can be anything between zero andmaximumvalue ofQ. * K. + Ky = Q & p. = py p2 2m.. + 2y y p 2m = Q p2 2m.. y 1 mm. . . .. . .. . . = Q K. 1 4 A 4 . . . . . . . . = Q . K. = A 4 A. . . . . . .Q C63: Find the kinetic energy of the .-particle emitted in the decay 238Pu 234U+ .. The atomic masses of 238Pu = 238.04955 a.m.u.; of 234U = 234.04095 a.m.u; of 4He = 4.002603 a.m.u. Neglect any recoilof the residualnucleus. Sol: Using energyconservation,

m(238Pu)c2 = m(234U)c2 + m(4He)c2 + K or K = [m(238Pu) � m(234U) � m(4He)]c2 = (238.04955 a.m.u. � 234.04095 a.m.u. � 4.002603 a.m.u.]c2 = 0.0059970 a.m.u. × (931.5)MeV/a.m.u. = 5.59MeV C64: Neon-23 beta decay in the followingway: 23 10 Ne 23 11 Na + 01e . + . Find theminimumandmaximumkinetic energy that the beta particle 01e . canhave. The atomicmasses of 23Ne and 23Na are 22.9945 u and 22.9898 u, respectively. Sol: Reactant Product

NUCLEAR PHYSICS www.physicsashok.in 33 23 10 Ne 22.9945 � 10me 23 11 Na 22.9898 � 11me 01e . me Total 22.9945 � 10 me Total 22.9898 � 10 me Mass defect, .m= 22.9945 u � 22.9898 u = 0.0047 u . Q = .m c2 = (0.0047 u) × 931.5MeV/u) = 4.4MeV The .-particle and neutrino share this energy.Themaximumkinetic energyofa beta particle inthis decayis, therefore, 4.4MeVwhenthe antineutrino does not get anyshare. Energyof . particle can range from0 to 4.4MeV. Endoergic collisions (a) IfQis negative, the reaction is endoergic. (b) For endoergic reaction to take place aminimumenergy has to be supplied. (c) Threshold energyEth: Theminimumamount of energy that a bombarding particle must have in order to initiate an endoergic reaction, is called threshold energyEth. Usingmomentumconservationalso,we get Eth = �Q 12 m 1 m . . . . . . . , wherem1 =mass of the bombarding particle,m2 =mass of the target nucleus. m1c2 + m2c2 + K1 = (m3 + m4)c2 + K3 + K4 & p1 = p3 + p4; Q + K1 = K3 + K4 Q + 21 1 p 2m = 23 3 p 2m + 24 4 p 2m = 23 3 p 2m + 2 1 3 4 (p p ) 2m.

. 2Q + 21 1 p 2m = 23 3 4 p 1 1 m m . . . . . . . � 1 3 4 2p p m + 214 pm 23 3 4 p 1 1 m m . . . . . . . � 1 3 4 2p p m + 21 p 4 1 1 1 m m . . . . . . . � 2Q = 0 2124 4p m � 4 3 4 1 1 m m . . . . . . . 2 1 4 1 p 1 1 m m . . . . . . . . . . . . .

� 2Q . 0 21 1 p 2m . � Q 3 4 3 4 1 m m m m m . . . ; K1 . �Q 3 4 3 4 1 m m m m m . . . If m1 + m2 ~� � Q 12 1 mm . . . . . . . C65: Howmuch energymust a bombarding proton possess to cause the reaction 73 Li + 11H 74Be + 10 n atomicmasses of 7Li, 1H, 7Be and 10 n are 7.01600 u, 1.0783 u, 7.01693 u and 1.0866 u respectively. Sol: Since themass of an atominclude themasses of the atomic electrons, the appropriate number of electron

NUCLEAR PHYSICS www.physicsashok.in 34 massesmust be subtracted fromthe given values to getmasses of nuclei. . Q-value, = [m( 73 Li ) � 3me+m(11H ) � me]c2 � [m(74Be ) � 4me + m(10 n )]c2 = [m( 73 Li ) +m(11H ) �m(74Be ) �m(10n )]c2 = (8.02383 � 8.02559)u.c2 = � 0.00176 u × (931.5MeV/c2) = �1.65MeV Negative sign ofQindicates endoergic reaction. Energymust be supplied for this reaction to take place. The energyis supplied as kinetic energy of the bombarding proton. The incident protonmust havemore than this energy because the system must posses some kinetic energy even after the reaction, so that momentumis conserved.Withmomentumconservationtakeninto account, theminimumkinetic energyof the incident can be foundwiththe formula. Eth = � 1 mM . . . . . . .Q = � 1 17 . . . . . . . (�1.65 MeV) = 1.89 MeV. C66.Making use ofthe table of atomicmasses find the energies of the following reachings Li7(., n) B10 Sol. 3Li7 + 2He4 . 5B10 + 0n1 Q = (7.01601 + 4.00260) � (10.0124 + 1.00867) Q = � 0.00300 amu = � 2.79 MeV Example 74: Find the energyof the reactionN14(., p) O17, if the kinetic energy of the incoming .-particle is T. = 4.0MeV and protonoutgoingat anangle .=60º to themotiondirection of the alpha-particle has a kinetic energyTp = 2.09MeV. N . = 60º . TO Tp p X .. T. Y Sol: T..= 4.0 MeV, Tp = 2.09MeV Let TO is kinetic energyafter collisionof oxygen. Reaction is, 7N14 + 2He4 8O17 + 1p1 Let Q - value of reactionQ. Energy conservation gives, Q + T. = TO + Tp .....(i) Momentumconservationalong x-directiongives, 2m T . . = O O 2m T .cos. + p p 2m T .cos. . 2m T . . � p p 2m T .cos.. = O O 2m T .cos. .....(ii)

Momentumconservationalong y-direction gives, p p 2m T .sin..= O O 2m T .sin. .....(iii) Squaring the equation (ii) and (iii) on both sides and adding the result,we get, ( 2m T . . � p p 2m T .cos.)2+ 2mpTp sin2. = 2mOTO . 2m.T. � 2 p p (2m T )(2m T ) . . .cos. + 2mpTp(cos2. + sin2.) = 2mOTO . TO = O 1 m [m. T. + mpTp � 2 p p m m T T . . .cos.] .....(iv) Fromequation (i), Q = TO + Tp � T. .....(v) Putting the value TO fromequation (iv) in equation (v),we get,

NUCLEAR PHYSICS www.physicsashok.in 35 Q = Tp � T. + O 1 m [m. T. + mp Tp � 2 p p m m T T . . . cos.] = 2.09 MeV � 4.0 MeV + 1 16 [1 × 4 + 1 × 2.09 � 2 4.1.4.2.09 .co60º] MeV . Q = �1.14MeV NUCLEAR FISSION Continuous research or artificialtransmuttionand especiallythe studyof inducedradioactivity, culminated in the discovery of nuclear fissionwhich is accompanied by the release of enormous amounts of energy. In ordinarynuclear disintegrations, both natural and artificial, the nucleus isonlychipped off rather than broken and accordingly, the amount of energyreleased is comparatively less i.e. fromabout 10 to 23MeV. It was discovered in 1939 that the heavyunstable uraniumnucleuswhen bombarded byneutrons splits into two almost equalfragmentswhich flyapart with great speed and the amount of energyreleased per fission is about 200MeV. This division of a nucleus into two approximately equalparts as called nuclear fission. Discovery of fission The starting point in the discovery of nuclear fission can be traced to the attempts of Fermi in 1934, to produce transuranic elements by bombarding uraniumwith neutrons. However, the fission process itself was discovered in 1939 by German radio chemists Otto Hahn and his two associates Meitner and Strassmann.After bombarding uraniumwith neutrons, theyperformed a series of chemical separations to identify the products. To their great surprise, they found that the atoms produced bythe bombardment of uraniumbelonged to elementswhich lie near the centre ofthe periodic table.Obviously, a uraniumnucleus after capturing neutronhad become so unstable that instead ofdisintegrating byejecting oneor two particles, it had split up into two parts. The actualfissionprocess canbeunderstoodwiththe helpoffigurewhich shows a uraniumnucleus capturing a neutron. 0n1 92U236 92p 143n 92U236 .-rays Unstable Antimony Nuclide 51Sb133 51p 82n .-rays Unstable

Niobium Nuclide 41p 48n 41Nb99 (a) (b) (c) The newly-formed nucleus of figure (b) isunstable and starts breaking up into two parts. Inbreaking up, the uraniumnucleus, behaving like a liquid drop, splashes out smalldroplets, i.e. neutron and .-rays. So great is the release of energy that the two fission fragments fly apart in opposite directions with tremendous speeds. It amy, however, be noted that not all uraniumnuclei break into Sb and Nb as shown in figure.

NUCLEAR PHYSICS www.physicsashok.in 36 There are at least 30 different ways inwhich a fissile nuclide can divide itself. The experimental evidence seems to favour pairs of fissionfragments of unequalmasses (asymmetrical fission) accompanied byone to five or some time more neutrons. In general, fission fragments are unstable nuclei containing an excess number of neutrons.After a series of .-emissions in which neutrons are converted into protons in the nucleus, a stable nuclide results. Out ofall the neutrons ejected during the fission of uranium, about 99 per cent are ejected inan extremely short interval oftime and are called prompt neutrons. The remaining one per cent of neutrons are emitted a little later and are called delayed neutrons. The delayed neutrons originate fromunstable fragments that decay by neutron emission on theirway to becoming stable nuclei. Itmaybe noted that divisionofa fissile nucleus into three fragments ofcomparable sizes (ternaryfission) has been observed although it is a rare event, occurring about 5 times permillionbinaryfissions. Types of Fission Reactions Historicallyspeaking, uraniumwas the first element to undergo fission.However, soon after itwas found that other elementsof high atomicweight could also bemade to undergo fissionand that particles other than neutron could be equallyeffective inthis respect. Naturaluraniumcontains three principal isotopeswith the following relative abundance: U238 99.28% 4.51 × 109 Y U235 0.714% 7.1 × 108 Y U234 0.006% 2.48 × 105 Y It is found that slowneutrons cause fission ofU235 but not ofmore abundant isotopeU238whichrequires fast neutronswith energies exceeding 1MeV. Similarly,Th232 and Pa231 undergo fissionwhenbombardedwith fast neutrons. Fission can also be produced in uraniumand thoriumby high-energy .-particles, protons, protons, deuterons and .-rays etc.Two other nuclideswhich do not occur in nature but have proved to be fissionalbe by neutrons of all energies are 92U233 and 94Pu239. In 1947, successful fission of bismuth, lead, thallium,mercury, gold, platinumand tantalumwas achieved inUSAbymeans of ..-particles, deutrons and neutrons of 100 MeV and more.With bismuth (Z = 83) fission was detected with 50MeV deuterons whereas tantalum(Z= 73) required .-particles of 400MeVenergy. It isworthnothing that onlythree fissilematerialsU233,U235, Pu239 are important inthe large-scale application of nuclear energy. Finally, some heavynucleihave beenfound to undergo spontaneous fission. In this process, nucleus divides in the ground statewithout bombardment byparticles fromoutside. Mass distribution of Fission Products During uraniumfission, a large number of nuclides of intermediate charge andmass are found. Their study is a promising source of information about the mechanismof the fission process itself and also offers the

possibility of discovering hitherto unknown nuclides. Investigations of the fission products ofU235 have shown that the range of their mass numbers is from72 to 158.About 97%of theU235 nuclei undergoing fission give fragmentswhich fall into two groups as shown inthe fission yield curve of figure . (i) light groupwithmass numbers from85 to 104 and (ii) heavygroupwithmass numbers from130 to 149. Themost probable type of fissionwhich occurs in about 7%of the total cases, gives fission productswith mass numbers 95 and 139.

NUCLEAR PHYSICS www.physicsashok.in 37 60 80 100 120 140 160 180 10 100 1000 10000 95 139 Mass Number (A) Number of Fragments Asmentioned earlier, fission fragments have toomany neutrons in their nuclei for stability. Consequently, most ofthemdecay byelectron emission. Each fragment starts a short radio-active series involvingmany emission of .-particles. These series are called fission decay series and chain has threemembers on the average althoughlonger and shorter chains occur frequently.One suchfission decaychain is shownin figure which startswith one of the unstable fragments of the fission ofU235 nucleus. 51 82 51Sb133 5m Unstable Antimony 52 81 52Te133 60m Unstable Tellurium O . 53 80 53I133 5d Unstable Iodine O .� 54 79 54Xe133 5d Unstable Xenon O .� 55 78 55Cs133 Stable Cesium O . Energy Distribution of Fission Products Energy distribution among the fission products can be found bymeasuring their kinetic energywith the help of suitable ionization chambers. The results of such studyonU235 fissionhave shown that the energy distribution curve is not uniform; rather it is a doublepeaked

curvewithmaxima at 67MeVand 100MeV. It is seen that while the greatest probability is for a fragment of 100MeV, the areas under the two peaks which represent the total number of particles in the two groups are approximatelyequal. 0 100 200 300 40040 60 80 100 120 Energy (MeV) Number of frangments 67 100 Neutron Emission in Nuclear Fission One ofthe notable features ofnuclear fission is thatwhile it is initiated byneutrons it is also accompanied by the emissionof fast-moving neutrons. The number of neutrons released depends on themode of fission and on the energyof the neutronswhichinduce fission. The average values for the number of neutrons emitted per thermalneutron absorbed bythe three important fissilematerials are given below: U235 2.43 U233 2.50 Pu239 2.89 These neutrons are emitted bythe fission fragments and not bythe compound nucleus.

NUCLEAR PHYSICS www.physicsashok.in 38 The neutrons emitted as a result of fission process (i.e. fissionneutrons) can be divided into two groups: (i) Prompt Neutrons: Thesemake up about 99.36%of the total fission neutrons and are ejected by the product nucleiwithin 10�14 second of the fission process. Prompt .-rays are also emitted at the same time. (ii) DelayedNeutrons: These constitute about 0.64%of the totalneutrons fromthe fission ofU235.These are emittedwithgraduallydecreasing intensityfor severalminutes after actual fissionprocess.Although the number ofdelayed neutrons is small, theyhave a strong influence on the time-dependent behaviour of chainreacting systems based on fission and play an important role in the controlofnuclear-fission reactors. Fissile and Fissionable Nuclides Elements likeU235, U233 and Pu239 undergo fission by neutrons of energy fromalmost zero upwards. Such nuclei are referred to as fissile nuclides. On the other hand, U238 and Th232 nuclei which have a fission threshold at 1MeV are said to be fissionable nuclides. In general, fissile nuclides have either an even number of protons and an odd number of neutrons or odd numbers ofboth. Fissionalbe nuclides, onthe other hand, have either even number ofprotons and neutrons or an odd number of protons and an even number of neutrons. Fission Energy One ofthe striking features ofthe fissionprocess is themagnitude of the energyreleasedwhich is about 200 MeV per fission ofU235 nuclide. Before 1939, the largest known nuclear reaction energywas 22.2MeV associatedwith Li6 (d, .) He4 reaction. The amount ofenergyreleased per fissionofU235 nuclidemaybe calculated bythe following threemethods: (i) Binding-energymethod:Asmentioned above all stable fission products havemass numbers in the range 72 to 158where the average binding energyper nucleon is about 8.5MeV.However, in the neighbourhood of uranium, its value is 7.6MeV. Hence, average binding energy per nucleon is (8.5 � 7.6) = 0.9MeV greater inthe fission products thanin the compound nucleus ofU235. The excess energyis released as fission energy. Its value is 235 × 0.9 ~. 200MeVper fission ofU235 nuclide (which has 235 nucleons). (ii) MassDefectMethod: The energy released per fission can also be estimated bycomparing themass of the interacting particles and the final fission products. As mentioned, U235 splits in manyways and the nuclei obtained in the greatest yield in fission by slow neutrons havemass numbers of 95 and 139. The fissionproducts being initiallyradioactive, undergo many .-emissions to formultimatelystable nuclides. Ifmolybdenum-95 and lanthanum-139 are takenas pair of stable products fromfission ofU235, the fission reaction can bewritten as 92U235 + 0n1 62Mo95 + 57La139 + 20n1 Comparingmasses on both sides of the above equationwe get, mass ofU235 nuclide = 235.124 amu

mass of one neutron = 1.009 amu Total = 236.133 amu mass ofMo95 nuclide = 94.946 amu mass of La139 nuclide = 138.955 amu mass oftwo neutrons = 2.018 amu Total = 235.919 amu mass of defect = 236.133 � 235.919 = 0.214 amu Therefore, energy released per fission ofU235 nucleus = 0.214 × 931 ~. 200 MeV (iii) Kinetic energymeasurementmethod: The total amount ofenergyreleased per fissionis equal to the sum ofthe following energies: (a) the kinetic energy of fission fragments.As seenfromfigure the average value of this energy for U235 is 167MeV.

NUCLEAR PHYSICS www.physicsashok.in 39 (b) the kinetic energyof fissionneutrons. Since the average number of neutrons emitted per fission ofU235 is 2.43 or say 2.5 and the average kinetic energyof these neutrons is 2MeV, total kinetic energyof fission neutrons is 2.5 × 2 = 5MeV. (c) the kinetic energy of prompt .-rays. Its value is about 7MeV. (d) the total energy of the decay process in the fission decay chains. This includes the energy carried away byradiations like .-rays, .-rays and neutrons. Its value is nearly 21 MeV. The totalof all the above energies is = 167 + 5 + 7 + 21 = 200 MeV C67:AU235 nucleus is fissioned bya thermalneutron and two fission fragments and two neutrons are produced. Compute the fissionenergyreleased if the average binding energyper nucleon is 7.8MeVin fissionedU235 nucleus and 8.6MeVin the fission fragments. Sol: Greater binding energyof the fissionfragments indicates that there has been release ofenergyduring fission of low-binding energynucleusU235. Fission energy released is = (234 × 8.6 � 236 × 7.8) = 171.6 MeV Theory of Fission Process The first attempt to explain themechanismoffission processwasmade byBohr andWheelerwho accounted formanyof the properties offission on the basis of the liquid-dropmodelof the nucleus. 1 2 3 4 5 6 The shape of the drop depends on a balance betweenthe surface tension forces and Coulombic repulsive forces. The excitation energy given to the drop during the capture of the slow or thermalneutron sets up oscillationswithin the drop. These oscillations tend to distort the spherical shape so that the drop becomes ellipsoid in shape.The surface tension forces trytomake the dropreturn to its originalspherical shapewhile the excitation energytends to distort the shape stillfurther. Ifthe excitation energyandhence oscillations are sufficientlylarge, the drop pattains the �dumbbell� shape as shownin figure. TheCoulombic repulsive forces thenpush the two �bells� further apart untilthe dumbbell splits into two similar drops eachofwhich assumes a spherical shape. The sequence of steps leading to fission is shown in figure. However, if the excitation energy is not large enough, the ellipsoidwillreturn to the sphericalshape. In that case, the excitation energyis givenout in the formof .-rays and the process becomes a radioactive capture process rather than fission process. LECTURE � 5 Nuclear Reactors : (a) Anuclear rectoris a systemdesignedto controlthe chainreactionoffissionwithcontinuous energyproduction. (b) Ausefulfactor for describing the levelof operation of a reactor is the reproduction constant K. It is defined as the average of neutrons available fromeach fission that will cause another fission. For a controlled or self sustained chain reactionKmust bemaintained close to unit.

i.e. K. 1 for controlled chain reaction.

NUCLEAR PHYSICS www.physicsashok.in 40 (c) Fuel: This is the fissionablematerial.Commonly usedmaterials areU238 enriched inU235 and plutonium(Pu239). (d) Moderator:Fastmoving neutrons cannot triggerthe fission ofU235 and have a very high chance of being captured by U238 which is not fissionable. It is therefore necessary to usemoderators to slowdown the neutrons. Control rods Shield Moderator Fuel elements Nuclear Reactor (e) Coolant: Air, water or CO2 are used as a coolant to remove the heat released inside the reactor. (f) Control Rods: Cd(cadmium) which is a good absorber of neutrons is used to controlthe rate of fissionand also to shut down the reactor in case of emergency. (g) Types of reactors: (i) Thermal reactors: In these reactors fission is produced by slowneutrons ro thermalneutrons. (ii) Breeder reactor: Breeder reactors generally use fast neutrons in these reactorsU238 is converted into Pu239 by capture of neutrons. Pu239 is fissionable. Thus such reactors also produce fuel in addition to the energyreleased through fission. (h) Critical mass: For a fuel there is a criticalmass belowwhich the fissionablematerial is completely safe. But for amass above the criticalmassmore neutrons are produced than are lost so that the chain reaction builds up rapidlyand the systemexplodes. The atomic bombare therefore stocked as subcriticalmass such that the combinedmass is greater than the criticalmass resulting in a spontaneous explosion. Example 75: In a nuclear reactor, fission is produced in 1 g for U235 (235.0439u) by a slowneutron (1.0087 u). Assume that 35Kr92 (91.8973 u) and 56Ba141 (140.9139 u) are produced inall reaction and no energy is lost. (a) Write the complete reaction, (b) Calculate the energy released per fission, (c) Calculate the total energy produced in kilowatt hour.Given 1 u = 931.5MeV/c2. Sol: (a) The nuclear fission reaction is 92U235 + 0n1 56Ba141 + 36K92 + 3 0n1 (b) Mass defect, .m= [(mu + mn) � (mBa + mKr + 3mn) .m= 256.0526 u � 235.8373 u = 0.2153 u Energyreleased per fission, Q = 0.2153 u × 931.5MeV/u = 200.6MeV (c) Number of atoms in 1g = 6.02 1023 235 . = 2.56 × 1021 Energy released in fission of 1 g ofU235 = 200.6 × 2.56 × 1021 MeV = 5.14 × 1023 MeV

= (5.14 × 1023) × (1.6 × 10�13)J = 8.2 × 1010 J = 8.2 × 1010 W-s = 10 6 8.2 10 3.6 10 .. KWh = 2.28 × 104 KWh

NUCLEAR PHYSICS www.physicsashok.in 41 Example 76: In neutron-induced binary fission of 92U235 (235.044) two stable end-products usuallyfound are 42Mo98 (97.905) and 54Xe136 (135.917).Assuming that these isotopes have come fromthe original fission process, find (i) what elementary particles are released (ii)mass defect of the reaction (iii) the equivalent energy released. Sol: (i) The reaction can bewritten as 0n1 + 93U235 = 43Mo98 + 51Xe136 It is seen that the totalZ-value of the two stable fission products is (42 + 54) =96. It is 4 unitsmore than that of 92U235. For balance, the original unstable fission products must have got Z = 92. Obviously, the originalunstable productsmust have emitted 4 .-particles before becoming stable.Now,mass number on right-hand side is 2 units less than on the left-hand side. It means that towfission neutronsmust have been produced.Hence, the fission reaction canbe represented by the following equation: 0n1 + 92U235 = 42Mo98 + 54Xe136 + 4 �1e0 + 20n1 (ii) .m= LHS mass � RHS mass LHSmass = (1.009 + 235.055) = 236.053 amu RHSmass = (97.905 + 135.917 + 4 × 0.0055 + 2 × 1.009) = 235.842 amu . .m = 236.053 � 235.842 = 0.211 amu (iii) Energy released = 0.211 × 931 = 196MeV Example 77. About 180MeVenergy is releasedwhen one nucleus of 92U235 undergoes fission. estimate the energy released from1 kg ofU235, assuming that each nucleus undergoes fission. Sol. 1 kg of U235 = 1000 g = 1000 235 mole . number of atoms = 1000 235 × 6.02 × 1023 . total energy released = 6.02 235 × 1026 × 180MeV = 6.02 18 235. × 1027 × 106 × 1.6 × 10�19 J = 7.37 × 1013 J C68. Calculate the energy released in slowneutron capture by Pu239.Mass of Pu239 = 239.127 amu, Pu240 = 240.1291 amu, 0n1 = 1.008665 amu. Sol. Energy released =mass defect in energy units = (239.127 + 1.008665 � 2401) amu = 0.006565 × 931MeV = 6.1 MeV Example 78.Anuclear reactor generates P =20MWpower at efficiency. =60%bynuclear fission of a radionuclidewhose half life isT= 2.2 years. If each fission releases energyE = 200MeV, calculate time during which µ = 10mole of the radionuclidewill be consumed completely. (Avogadro number, N = 6 × 1023, loge 2 = 0.693, 1 year = 3.15 × 107 s)

NUCLEAR PHYSICS www.physicsashok.in 42 Sol. To operate the nuclear reactor, let the number of fissions required per second be n0. Thenenergy released per second byfission reactions = n0E Since, efficiencyofthe reactor is ., therefore, power output fromthe reactor = .n0E. But it is equal to P therefore, P = .n0E or 0 n PE . . Let at an instant number of nuclei of radionuclide be n then rate of decay= .nwhere . is decay constant which is equal to e log 2 T . Hence, net rate of decrease of nuclei 0 dn n n dt . . . . . . .. .. or e e dn n log 2 P n Elog 2 PT dt T E ET . . . . . .. . . . . . . . . . e dn dt PT n E log 2 Et . . . . . ...(1) At t = 0, number of nuclei are n = µNand time t is to be calculated when all the nuclei are consumed or when n = 0, t = ? Integrating equation (1)withthese limits, 0 t µN 0 e dn dt PT n E log 2 ET . . . . . . . . e 8 . . e e e T µN Elog 2 t log 1 10 log 1.0576 s log 2 PT . . . . . . . . . . Example 79. The element Curium 248 96 Cmhas a mean life of . = 1013 second. Its primary decaymodes are spontaneous fission and .-decay, the former with a probabilityof P1 = 8%and latter with a probability of P2 = 92%. each fission releasesE = 200MeVenergy. The masses involved in .-decay are as follows 248 96 Cm= 248.072220 u, 244 94 Pu = 244.06400 u and 42He = 4.002603 u.

Calculate the power output froma sample ofN = 1020 Cmatoms. (1 u = 931MeV/c2) Sol. Decayconstant, 1 10 13 sec 1 mean life ( ) . . . . . . Rate of decay froma sample of N atom,A= .N = 107 sec�1 Since, probabilities of fission and .-decay are P1 and P2 respectively, therefore, rate of decay due to fission, A1 = P1A or A1 = 8 × 105 sec�1 and rate of decay due to .-emission,A2 = P2A= 9.2 × 106 sec�1. Since, each fission releases energyE, therefore, rate of release of energydue to fission =A1 . E

NUCLEAR PHYSICS www.physicsashok.in 43 Equation of .-decay is 248 244 4 96 94 2 Cm... Pu . He Massmost during each .-decay, ..m= [248.072220 � (244.06400 + 4.002603)] u = 5.617 × 10�3 u . Energy released during each .-decay, E´ = 5.617 ×10�3 × 931 MeV E´ = 5.23MeV . Rate of release of energy due to .-decay =A2 . E´ . Total rate of release of energy =A1E +A2E´ But totalrate of release of energy is equal to power output. Therefore, power output, P = A1E + A2E´ = 3.33 × 10�5W Nuclear fusion It is the process of combining or fusing two lighter nuclei into a stable and heavier nuclide. Inthis case also, large amount of energyis released becausemass of the product nucleus is less than themasses oftwo nuclei which are fused. Many reactions between nuclei of lowmass numbers have been brought about by accelerating one or the other nucleus in a suitable manner. These are often fusion processes accompanied by release of energy. However, reactions involving artificially-accelerated particles cannot be regarded as ofmuch significance for the utilizationof nuclear energy.To have practicalvalue, fusion reactionsmust occur in suchamanner as tomake themself-sustaining, i.e.more energymust be released thanis consumed in initiating the reaction. It is thought that the energyliberated inthe Sun and other stars of themainsequence type is due to the nuclear fusion reactions occurring at the very high stellar temperature of 30million ºK. Suchprocesses are called thermonuclear reactions because they are temperature-dependent. Steller Thermonuclear Reactions: Following two sets of thermonuclear reactions have been proposed as sources of energy in the Sun and other stars of themain sequence: (i) proton-proton (p - p) chain and (ii) carbon-nitrogen (C-N) cycle. At lowtemperatures corresponding to those in the Sunwhenit was first formed, the proton-proton chain was predominant. In the present state of the Sun with its higher central temperature and larger He4 concentration, the C-Ncycle is supposed to be the main source of its energy. Proton-Proton Chain It is so called because the step involves the combination of two protons.When two protons fuse, they produce a deuteronnucleus, a positron and a neutrino thus: 1H1 + 1H1 = 1D2 + 1e0 + v × 2 The deutron thencombineswith another proton to yield helium-3. 1D2 + 1H1 = 2He2 + . × 2 The two helium-3 nuclei fuse to produce helium-4 thus 2He3 + 2He3 = 2He4 + 21H1 + 1e0 It should be noted that for the third reaction to occur, each of the first two reactionsmust occur twice. The net effect ofthe reactions is 41H1 = 2He4 + 21e0 + 2. + 2v

NUCLEAR PHYSICS www.physicsashok.in 44 Obviously, four hydrogen atoms are fused to produce one heliumatomwith a total energyrelease of about 26.7 MeV.When the kinetic energy of neutrions is substracted, the energy is 26.2 MeV. The emitted positrons are annihilated by free electronswith the production of .-rays. Carbon-Nitrogen Cycle It was proposed byH.A. Bethe to account for the energy production in the Sun and other stars ofmain sequence. In this cycle, carbon acts as a nuclear catalyst. The cycle startswhen a proton (hydrogen atom) first interactswith carbon-12 nucleuswiththe release of fusionenergythus 6C12 + 1H1 = 7N13 + . The product N13 is known to be radioactive, emitting a positronwith a half-life of 10minutes. Hence, it decays in a very short time according to the relation 7N13 = 6C13 + 1e0 + v The stable C13 nucleus reactswith another proton, therebyliberatingmore energy 6C13 + 1H1 = 7N14 + . The stable product N14 combineswith third proton thus 7N14 + 1H1 = 8O15 + . TheO15 nucleus is a positron emitterwith a half-life of 2.06minuteswhich decays by the process 8O15 = 7N15 + 1e0 + v Finally, the resultingN15 nucleus interactswiththe fourthproton thus: 7N15 + 1H1 = 6C12 + 2He4 By adding up the above six equations and cancelling out those nucleiwhich appear on both sides, it is seen that four hydrogen atoms are consumed and, in return, 2 positrons, 3 .-rays and one heliumnucleus are created. In otherwords, hydrogen is burned and heliumis created. The overall processmay bewritten as 41H1 = 2He4 + 21e0 + 24.7 MeV The annihilation of positrons supplies an additional energy of 2MeVso that total energyreleased is 26.7 MeV. The fusion energy releasedmayalso be found by the loss ofmass during the above reaction: 41H1 = 4 × 1.008144 = 4.032576 amu 2He4 = 4.003873 amu 21e0 = 2 × 0.000558 = 0.001115 amu . mass loss = 4.032576 � (4.003873 + 0.001115) or .m = 0.028857 amu . energy released = 931 × 0.028857 = 26.7MeV It is worthnoting that the above energy release is less than that in nuclear fission. However, its value is 26.7/4 = 6.7MeVper nucleon as compared to less than 1MeVper nucleon in fission process. Controlled Thermonuclear Reactions The fact that nuclear fusion reactions release large amounts of energy, as in stars, has attracted much attention and continuous search is beingmade for finding practicalmeans or controlled release of such energy. It has however, been found that reactions ofC-Ncycle and proton-proton chain occur too slowly to be of any practical use.Other thermonuclear reactionswhich occurmuchmore rapidlyand depend on aboundant hydrogen isotopes like deuteron (1D2 or 1H2) and tritium(1T3 or 1H3) a

nd hence seemmore practicalproposition, are as under: (i) 1H2 + 1H2 = 2He3 + 0n1 + 3.3 MeV

NUCLEAR PHYSICS NUCLEAR PHYSICS or 1D2+ 1D2 = 2He3+ 0n1+3.3MeV

(ii) 1H2+ 1H2 = 1H3+ 1H1+4MeV or 1D2+ 1D2 = 1T3+ 1H1+4MeV (iii) 1H2 + 1H3 = 2He4 + 0n1 + 17.6 MeV or 1D2+ 1T3 = 2He4+ 0n1+17.6MeV The most important aspect of the above nuclear fusion reactions is that deuterium is available easily and aboundantly. It occurs in nature with an aboundance of one part in six thousands of hydrogen and can be separated fromthe lighter isotope quite cheaply. Five litres of water contain about 1/8 gram of deuterium but its energy content if it could be used as a fuel in a thermo-nuclear reactor, would be equivalent to 130 litres of petrol! The more than 5 × 1019 kg of water present in the oceans could thus supply world�s power requirement for severalmillion years at negligible cost ifthe deuteriumcould be utilized to provide energyby fusion reactions. However, as discussed below, there are some difficult problems to be solved before man-made controlled fusion reactors can become a reality. Condition for Controlled Fusion In order to provide useful energy, the fusion process must be self-sustaining. Once the temperature of deuterium (or a mixture of D2 and T3) has been raised to the point at which fusion occurs at an appreciable rate, the energyreleased must be sufficient, at least, to maintain that temperature. The minimumtemperature is known as critical ignition temperature and may be defined as that temperature above which the rate of energy production by fusion exceeds the rate of energy loss. Its value is about 5 keV (i.e. 50 million ºK) for a D-D-reaction. At these temperatures, the atoms are entirely stripped of their electrons. The result is a completely ionized gas or plasma consisting of atomic nuclei (like deuterons, tritons and protons) and electrons in rapid random motion. It is practically impossible to contain such plasma in walls of ordinary materials. In the Sun, the fusion reactions are contained bya tremendous gravitational pressure. Such a high pressure is yet not available for controlled thermo-nuclear reactions on earth although the plasma can be contained in a magnetic field. Hence, main problem is to devise an apparatus in which plasma can be obtained by meansofa magnetic field at the kinetic temperaturesrequired for the fusionreactions to proceed. Another necessary condition for a self-sustaining thermonuclear system is known as Lawson criterion. It is based on the requirement that in the operation of a fusion reactor, the total usefulrecoverable energyshould be at least sufficient to maintain the temperature of te reacting nuclei. Lawson criterion can be expressed in terms of the product �nt� where nis the number of reacting nuclei per m3 and t is the time inseconds in which the thermonuclear reaction takes place. The minimum value of �nt� for D-T system is 7 × 1019 and D-D system is 2 × 1021. In controlled thermonuclear reactions, t is taken as the time during which the high-

temperature plasma can be confined. It will be seen from above that both the critical ignition temperature and Lawson criterion are much more favourablefor D-TsystemthanforD-Dsystem. Buttheformersystemhasthedraw-backthat itrequires tritium which has to be obtained by nuclear reactions (because it does not occur in nature). Tritium can, however, bemadebybombardinglithiumwithslowneutronsinareactorthus:

3Li6+ 0n1= 1H3+ 2He4+4.8MeV Thisreactioncanbemadetoservetwousefulpurposes.InthethermonuclearreactorusingD-T reaction, the escaping neutrons carry off much of the energy (about 14 MeV per neutron). This energy can be converted into heat byslowing down these fast neutrons in a blanket of berylliumsurrounding the reactor in whichplasmaisproduced. Theslowneutronsarecapturedbylithiumwhichproducestritium.Theblanket couldthusconsistofmoderator (i.e.beryllium),coolantandlithium.Theheatgeneratedbythemoderation and absorption of neutrons could thus be transferred by coolant to external heat exchangers and then to turbines whichcould run alternators.

www.physicsashok.in 45

NUCLEAR PHYSICS www.physicsashok.in 46 Hydrogen bomb This bomb is 1000 timesmore powerfulthan the atomic bombwhich is based onnuclear fission.Hydrogen bomb is based on the fusion if the hydrogen atoms into heavier ones by the thermonuclear reactionswith release ofenormous energy.The essentialconditions for the operationof the hydrogen bombare extremely high temperatures and pressures required for the fusion to start.Once started, the fusionitselfmaintains the temperatures to keep the process going. For this purpose, the atombomb (fission bomb) is used as a primer which, byfirst exploding, provides the high temperature and pressure necessaryfor the successful working of the hydrogen bomb (fusion bomb). Afusion bomb is superior to a fission bomb because of the following reasons: (i) The energy release in a hydrogen bomb is open-ended i.e. it has no upper limit. It depends on howmuch fusiblematerial is present in the bomb. (ii) It hasno limitation of a critical size of the fusiblematerialunlike anatomic bomb. If the activematerial in an atomic bombexceeds the critical size, spontaneous explosion results.Hydrogen bomb cannot explode unless �ignited� i.e., heated to critical ignition temperature and anyamount of fusiblematerial is safe until ignited. Thus the amount of fusiblematerialin a hydrogen bombis not limited. Cobalt Bomb It consists of a hydrogen bomb which is encased in a sheath of metallic cobalt and is more lethal and destructive thana simple uncased hydrogenbomb.When the hydrogenbomb explodes, it gives off neutrons which act on the cobalt cover and render it intensivelyradioactive due to the formation ofCo60 that is 300 timesmore powerfulthan radium.During explosion, the radioactive cobalt is pulverised and converted into a gigantic radioactive cloudwhich canspread over thousands ofkilometres killing everything living in that area. Fission and Fusion: One thing common between the two nuclear processes si that they release very large amounts of energy. But there aremanydifferences inthemechanisms of the two processes. (i) Fission involves breaking up of a heavy nucleus into lighter nuclei. Fusion, on the other hand, involves combining oftwo lighter nuclei into on heavynucleus. (ii) The links of the fission process are neutronswhile the links of a fusion process are protons. (iii) Fission proceeds best with thermalneutronswhere thermalmeans roomtemperature. Fusion proceeds best withthermalparticleswhere thermalmeans temperatures ofmillions of ºK. Example 80: the masses of 1H1 and 2He4 atoms are 1.00813 amu and 4.00386 amu respectively. Howmuch hydrogenmust be converted to heliumif solar constant is 1.35 kW/m2 anthe earth is 1.5 × 1011mfromthe Sun. Sol: This thermo-nuclear reactionmay bewritten, inits essentials as 41H1 = 2He4 + 1e0 Neglecting the two positrons, it is seen that 4 hydrogen atoms fuse to produce o

ne atomof helium. Mass of 4 hydrogen atoms = 4 × 1.00813 = 4.03252 amu Mass ofone heliumatom = 4.00386 amu Decrease inmass, .m = 4.03252 � 4.00386 = 0.002866 amu energy produced = 0.02866 × 931 = 26.68 MeV This is the energy released when four hydrogen atoms fuse. Hence, energy produced by one hydrogen atom. = 26.68 4 = 6.67 MeV = 6.67 × 1.6 × 10�13 = 10.67 × 10�13J = 6.02 × 1026 × 10.67 × 10�13 = 6.42 × 1014 J

NUCLEAR PHYSICS www.physicsashok.in 47 The solar constant represents the amount of energy received per second by 1m2 area held perpendicular to the Sun�s rays at a distance equal to the mean distance of the earth fromthe Sun. Sun 1.5 × 1011 Given value of solar constant = 1.35 kW/m2 = 1.35 × 1000 W/m3 = 1350 J/s/m2 Total energyemitted bythe Sunis equal to the energyreceived bythe inner surface of the imaginary sphere drawnwith Sun as centre and radius = 1.5 × 1011m. Surface area of the sphere = 4.R2 = 4. × (1.5 × 1011)2 = 28.28 × 1022 m2 . energy received by this surface area per second is = 1350 × 28.28 × 1022 = 38.18 × 1025 J/s It also represents the energy emitted by the Sun per second. Mass of hydrogen consumed is = 25 14 38.18 10 6.42 10 . . = 5.59 × 1011 kg/s = 5.95 × 108 tones/second C69: Calculate the energyliberatedwhen aHeliumnucleus is formed bythe fusion of two deuteriumnuclei. The mass 1H2 = 2.01478 a.m.u. and mass of 2He4 = 4.00388 amu. Sol: The reactionmaybewritten as 1H2 + 1H2 = 2He4 + Q . 2.01478 + 2.01478 = 4.00388 + Q , . Q = 0.02568 amu = 0.02568 × 931 = 23.9 MeV C70. In the fusion reaction 1H2 + 1H2 . 2He3 + 0n1, deuteron, heliumand the neutron havemasses 2.015 amu, 3.017 amu and 1.009 amu, respectively.Estimate the total energyreleased if 1 kg of deuteriumundergoes complete fusion. Sol. Mass difference = 2 × 2.015 � (3.017 + 1.009) = 0.004 amu . energy released = 0.004 × 931MeV = 3.724MeV Energy released per deuteron = 12 × 3.724MeV No. of deuterons in 1 kg = 6.02 1026 2. . energy released per kg = 1.862 × 3.01 × 1026 MeV = 5.6 × 1026 × 106 × 1.6 × 10�19 . 9 × 1013 C71. In some stars, three 2He4 nuclides fuse together to form6C12 ofmass 12.0000 amu. Howmuch energy is released per fusion of 6C12 ? Rest mass of 2He4 = 4.002603 amu. Find also the rate of consumption of heliumto maintain the radiative power of the star at 4 × 1021MW.mass of 2He4 atom= 6.9 × 10�27 kg. Sol. Energy released per fusion =mass defect in amu = (3 × 4.00263 � 12.0000) amu = 0.007809 × 931MeV = 7.27 MeV . energy released by n atoms7.27 1.6 10 19 106 n J

3 . . . . . . = 4 × 1021 × 106 (given) , n = 1.103163 × 1040 . mass of heliumatoms burnt per second = 1.03163 × 1040 × 6.9 × 10�27 kg = 7.1 × 1013 kg