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### Transcript of Physics 2203, Fall 2012 Modern Physics - 2203, Fall 2012 Modern Physics . Friday, Aug. 31st , 2012:

• Physics2203,Fall2012ModernPhysics

.

Friday,Aug.31st,2012:FinishCh.2

Announcements:Tutorialsession4:30pmTuesdayinNicholson102QuiztodayFallbreakcanceled!MondayisLaborDay.

• E2 = pc( )2 + mc2( )2

EK = m0c2 1( ) = m0c2 m0c2

E = m0c2 =

m0c2

1 u2

c2

= m0c2 EK +m0c

2

EK = p2c2 + mc2( )2 mc2

=vcpcE E = m0c

2 = Mc2

M = m0

• An electron volt (eV) is the energy to move an electron though one volt.

1.0eV = e 1.0V( ) = 1.602x1019C( ) 1.0V( ) =1.602x1019 J

J = 1eV1.602x1019C( )

EnergyismeasuredinJoules.ConvertJtoeV

Example : Rest energy of an electron

m0c2 (electron) =

8.19x1015 J( )eV1.602x1019 J

= 5.11x105eV = 0.511MeV

m0c2 (electron) = 0.511MeV

m0c2 (neutron) = 939.57MeV

m0c2 (protron) = 938.3MeV

Units for mass

m0 MeVc2

m0 (electron) =0.511MeV

c2

• Units for momentum: E2 = pc( )2 + m0c( )2

pc Energy eV ,KeV ,MeV

pc = m0vc1 v2 / c2

Periodic Table notation: ZAP

P The chemical element --H, He, Ar, etc.A=Z+N Z # protrons, N # neutrons

Atomic Mass unit u, defined as 112

mass of 612C

1u = 931.494013Mev / c2

Uranium 238U 238.0507u

• E = mc2 =mc2

1 u2

c2

=E0

1 u2

c2

= EK + E0Asimpleexperiment:twoblocksofwoodwithequalmassmandkineMcenergyK,aremovingtowardeachotherwithvelocityv.Aspringplacedbetweenthemiscompressedandlocksinplaceastheycollide.LetslookattheconservaMonofmassenergy

Mass Energy before: E=2mc2 + 2KMass Energy after: E=Mc2

Since Energy is conserved we haveE=2mc2 + 2K=Mc2

M is greater than 2m because K went into mass

M = M 2m = 2Kc2

fr =M 2m

2m=

Kmc2

for real systems ~1016

HW:Whatisthemassofacompressedspring?

In Relativistic Mechanics Momentum and Total Energy are conserved!

• ExamplesofEnergytoMassExchangeIonizaMon

ChemicalBindingEnergy

Fusion

Fission

NuclearreacMons:

H p + e : 13.6 eV: M=13.6eVc2

H2O 2H +O : 3 eV: M=3eVc2

232Th 228Ra + 4He: M= 4MeVc2

12H + 1

2H 24He+ energy: M= 23.9MeV

c2

p + p p + p + p + p : antiprotron

• TheanMprotronwasdiscoveredin1956throughthefollowingreacMon

p

p + p p + p + p + p

FindtheminimumkineMcenergyoftheacceleratedprotoninthefigure.ThisiscalledthethresholdkineMcenergy,forwhichthefinalpar9clesmovetogetherasiftheywereasingleunit.

Conservation of EnergyE p +mpc

2 = 4E 'p

Conservation of Momentump p = 4 p 'p

Here E'pand p'p arefor each of the 4 particles

Ep2 mpc

2( ) = 4 E 'p2 mpc2( )E 'p =E p +mpc

2

4Ep2 mpc

2( ) = Ep +mpc2( )216 mpc

2( )

Ep = 7mpc2

K = Ep mpc2 = 6mpc

2 = 6 938MeV( ) = 5628Mev = 5.628GeV

E2 = pc( )2 + mc2( )2

• AneutralKmeson(mass497.7MeV/c2)ismovingwithakineMcenergyof77.0Mev.Itdecaysintoameson(mass139.6MeV/c2)andanotherparMcle(A)ofunknownmass.ThemesonismovinginthedirecMonoftheoriginalKmesonwithamomentum381.6MeV/c.(a)FindthemomentumandtotalenergyoftheunknownparMcle.(b)FindthemassoftheunknownparMcle.

K + ATotal Energy and Momentum of K meson areEK = KK +mKc

2 = 77.0Mev + 497.7 MeV=574.7 Mev

pK =1c

EK2 + mKc

2( )2 = 287.4Mev / c

Total Energy of meson is

E = cp( )2+ mc

2( )2 = 381.6MeV( )2 + 139.6MeV( )2 = 406.3MeVConservation of momentum requires pK = p + pApA = pK p = 287.4 381.6( )MeV / c = 94.2Mev / cConservation of Energy requires EK = E + EAEA = EK E = 574.7 406.3( )MeV =168.4Mev

(b) find mass: mAc2 = EA

2 + cpA( )2= 168.4( )2 + 94.2( )2MeV =139.6MeV

• BindingEnergyoftheHydrogenAtom:Thebindingenergiesofelectronstothenucleiofatomsaremuchsmallerthannuclearbindingenergies.Thebindingenergyforanelectrontoaproton(Bohrmodel)is13.6eV.Howmuchmassislostwhenanelectronandprotonfromahydrogenatom?

mc2 = E(binding) = 13.6eVmHc

2 mpc2 = 938.3MeV

mmH

1.4x108

Thisisagainabeforeandaberproblem:Beforeyouhaveanisolatedelectronandproton.ACeryouhaveaHatomwhosebindingenergyis13.6eV.

• (a) Howmuchlighterisamoleculeofwaterthantwohydrogenatomsandanoxygenatom?Thebindingenergyofwateris~3eV.

(b) FindthefracMonallossofmasspergramofwaterformed.(c) Findthetotalenergyreleased(mainlyasheatandlight)when1gramofwaterisformed?

(a) M= mH + mH + mO( ) MH2O =E(binding)

c2

M=3.0eV( ) 1.6x1019 J / eV( )

3.0x108m / s( )2= 5.3x1036 kg Reallysmall!

(b) MMH2O

=E(binding)MH2Oc

2

MH2O =18u : u is 1/12 of the mass of Carbon (6 protron, 6 neutrons)

MH2O =18 1.66x1027 kg( )

MMH2O

= 5.3x1036 kg

18 1.66x1027 kg( )2=1.8x1010 SMllsmall!

(c) E=mc2 = 1.8x1010( ) 103kg( ) 3x108m / s( )2 = 16kJ Big!

• Fusion:Energyisgainedbytakingtwolightatomsandcombiningthemintoanatomwithaheaviernucleilaterthissemester.Forexample:

12H + 1

2H = 24He + Energy

Ec2= mass 1

2H + 12H( ) mass 24He( )

E = 3751.226 3727.379( )MeVE = 23.9MeV

• FissionReac9ons:ThedecayofaheavyradioacMvenucleusatrestintoseverallighterparMclesemieedwithlargekineMcenergiesisagreatexampleofmassenergyconversion.AnucleusofmassMundergoesfissionintoparMcleswithmassesM1,M2,andM3,withspeedsofu1,u2,andu3.

The conservation or Relativistic Energy Requires that

Mc2 = M1c2

1 u12

c2

+M2c

2

1 u22

c2

+M3c

2

1 u32

c2

This is a very important equation to remember

The Equation above is true if M > M1 + M2 + M3( )Disintegration Energy Q defined

Q= M M1 + M2 + M3( ) c2

Example in our text (pg 61)232Th 228Ra + 4HeThe offspring have 4 Mev Kinetic Energy

AppendixD:1u=1.66x1027kg:

M = 0.004u( )1.7x1027 kg( )

uM = 7x1030 kgMc2 = 6.3x1013J = 4MeV

• ChangeintheSolarMass:ComputetherateatwhichtheSunislosingmass,giventhatthemeanradiusRoftheEarthsorbitis1.50x108kmandtheintensityofthesolarradiaMonontheEarthis1.36x103W/m2(calledsolarconstant).

Assume that the sun radiates uniformly as a sphere of radius RP= area of shere( ) solar constant( )P= 4R2( ) 1.36x103W / m2( )P=4 1.50x1011m( )2 1.36x103W / m2( )P = 3.85x1026 J / s

m= E(lost)c2

m = 3.85x1026 J / s

3x108m / s( )2= 4.3x109 kg / s

Thisisabout4millionmetrictons

Ifthisrateofmasslossremainsconstantandwithafusionmasstoenergyconversionefficiencyofabout1percent,theSunspresentmassof~2.0x1030kgwillonlylastforabout1011moreyears!

• InClassicalMechanicswecanthaveaparMclewithm=0,becauseEandparezero.

InRela9vis9cMechanicswecanhaveaparMclewithm=0

E = m0c2

p= m0u

E2 = pc( )2 + m0c2( )

2

=vcpcE

NowlookatwhathappenstotheequaMonsintherightboxifm=0.

E = pc =1

NowlookatwhathappenstotheequaMonsinthelebboxifm=0.

E = m0c2 0

p= m0u

0

Undefined

• InRela9vis9cMechanicswecanhaveaparMclewithm=0.Itisaphoton.Chapter4

AphotonisaconsequenceofQuantumMechanics(Einstein).ItistheparMclenatureorlightwaves.Theenergycomesintermsofphotonseachwithadiscreteenergyandmomentumdependinguponthewavelengthofthelight.

Youmusthavedonetheexperimentwithasupportedplatewithonesidepolishedandtheothersideblack.Whenlightshinesonthistheplaterotatesbecauseofthemomentumchangefromthephotons.

Our example of ionization of H + H e+ p

PhotonsaremasslessparMclesv=c:E=pcNeutrinoswerebelievedtobemassless,butrecentexperimentsshowm~105me.

Gravitonsarerelatedtogravitythewayphotonsaretolightshouldhavem=0.NoexperimentalevidenceLIGO

• Quiz

• ConsidertheinelasMccollisionoftwoequalmassobjectsshowninthefigure.

NowconsidertheSsystemmovingwithobject#1atavelocityofv.

Provethatbyusingthenewdefini9onofmomentumthatmomentumisconservedintheScoordinatesystem.

p=

mu

1 u2

c2

u 'x =ux v

1 vuxc2

we need to find v2 'and V' using the transform

v2 ' =v2 v

1 v2vc2

=v v

1 v2

c2

=2v

1 v2

c2

V ' = V v

1 Vvc2

=0 v

1 0( )vc2

= v