Physics2203,Fall2012ModernPhysics

.

 Friday,Aug.31st,2012:FinishCh.2

 Announcements:• Tutorialsession4:30pmTuesdayinNicholson102• Quiztoday• Fallbreakcanceled!• MondayisLaborDay.

E2 = pc( )2 + mc2( )2

EK = m0c2 γ −1( ) = γm0c

2 −m0c2

E = γm0c2 =

m0c2

1− u2

c2

= γm0c2 ≡ EK +m0c

2

EK = p2c2 + mc2( )2 −mc2

β =vc≡pcE E = γm0c

2 = Mc2

M = γm0

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

1.0eV = e 1.0V( ) = 1.602x10−19C( ) 1.0V( ) =1.602x10−19 J

J = 1eV1.602x10−19C( )

EnergyismeasuredinJoules.ConvertJtoeV

Example : Rest energy of an electron

m0c2 (electron) =

8.19x10−15 J( )eV1.602x10−19 J

= 5.11x105eV = 0.511MeV

m0c2 (electron) = 0.511MeV

m0c2 (neutron) = 939.57MeV

m0c2 (protron) = 938.3MeV

Units for mass

m0 →MeVc2

m0 (electron) = 0.511MeVc2

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 = γmc2 =mc2

1− u2

c2

=E0

1− u2

c2

= EK + E0Asimpleexperiment:twoblocksofwoodwithequalmassmandkineMcenergyK,aremovingtowardeachotherwithvelocityv.Aspringplacedbetweenthemiscompressedandlocksinplaceastheycollide.LetslookattheconservaMonofmass‐energy

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

ΔM = M − 2m =2Kc2

fr =M − 2m

2m=

Kmc2

for real systems ~10−16

HW:Whatisthemassofacompressedspring?

In Relativistic Mechanics Momentum and Total Energy are conserved!

ExamplesofEnergytoMassExchange• IonizaMon

• ChemicalBindingEnergy

• Fusion

• Fission

• NuclearreacMons:

H → p + e : 13.6 eV: ΔM=13.6eVc2

H2O→ 2H +O : 3 eV: ΔM= 3eVc2

232Th→ 228Ra + 4He: ΔM= 4MeVc2

12H + 1

2H → 24He+ energy: ΔM= 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− mpc

2( )E 'p =E p +mpc

2

4Ep2 − mpc

2( ) = Ep +mpc2( )2 −16 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.Itdecaysintoaπmeson(mass139.6MeV/c2)andanotherparMcle(A)ofunknownmass.TheπmesonismovinginthedirecMonoftheoriginalKmesonwithamomentum381.6MeV/c.(a)FindthemomentumandtotalenergyoftheunknownparMcle.(b)FindthemassoftheunknownparMcle.

K →π + 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 π meson is

Eπ = cpπ( )2+ mπc

2( )2= 381.6MeV( )2

+ 139.6MeV( )2= 406.3MeV

Conservation of momentum requires pK = pπ + pApA = pK − pπ = 287.4 − 381.6( )MeV / c = −94.2Mev / c

Conservation of Energy requires EK = Eπ + EA

EA = EK − Eπ = 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?

Δmc2 = E(binding) = 13.6eVmHc

2 ≈ mpc2 = 938.3MeV

ΔmmH

≈ 1.4x10−8

Thisisagainabeforeandaberproblem:Beforeyouhaveanisolatedelectronandproton.ACeryouhaveaHatomwhosebindingenergyis13.6eV.

(a)  Howmuchlighterisamoleculeofwaterthantwohydrogenatomsandanoxygenatom?Thebindingenergyofwateris~3eV.

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

(a) ΔM= mH + mH + mO( ) − MH2O=E(binding)

c2

ΔM=3.0eV( ) 1.6x10−19 J / eV( )

3.0x108m / s( )2 = 5.3x10−36 kg Reallysmall!

(b) ΔMMH2O

=E(binding)MH2O

c2

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

MH2O=18 1.66x10−27 kg( )

ΔMMH2O

= 5.3x10−36 kg18 1.66x10−27 kg( )2 =1.8x10−10 SMllsmall!

(c) E=Δmc2 = 1.8x10−10( ) 10−3kg( ) 3x108m / s( )2= 16kJ Big!

Fusion:Energyisgainedbytakingtwolightatomsandcombiningthemintoanatomwithaheaviernuclei‐laterthissemester.Forexample:

12H + 1

2H = 24He + Energy

Ec2= mass 1

2H + 12H( ) − mass 2

4He( )E = 3751.226 − 3727.379( )MeVE = 23.9MeV

FissionReac9ons:ThedecayofaheavyradioacMvenucleusatrestintoseverallighterparMclesemieedwithlargekineMcenergiesisagreatexampleofmass‐energyconversion.AnucleusofmassMundergoesfissionintoparMcleswithmassesM1,M2,andM3,withspeedsofu1,u2,andu3.

The conservation or Relativistic Energy Requires that

Mc2 =M1c

2

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=1.66x10‐27kg:

ΔM = 0.004u( )1.7x10−27 kg( )

uΔM = 7x10−30 kgΔMc2 = 6.3x10−13J = 4MeV

ChangeintheSolarMass:ComputetherateatwhichtheSunislosingmass,giventhatthemeanradiusRoftheEarth’sorbitis1.50x108kmandtheintensityofthesolarradiaMonontheEarthis1.36x103W/m2(calledsolarconstant).

Assume that the sun radiates uniformly as a sphere of radius RP= area of shere( ) solar constant( )P= 4πR2( ) 1.36x103W / m2( )P=4π 1.50x1011m( )2

1.36x103W / m2( )P = 3.85x1026 J / s

m= E(lost)c2

m =3.85x1026 J / s3x108m / s( )2

= 4.3x109 kg / s

Thisisabout4millionmetrictons

Ifthisrateofmasslossremainsconstantandwithafusionmass‐to‐energyconversionefficiencyofabout1percent,theSun’spresentmassof~2.0x1030kgwill“only”lastforabout1011moreyears!

InClassicalMechanicswecan’thaveaparMclewithm=0,becauseEandparezero.

InRela9vis9cMechanicswecanhaveaparMclewithm=0

E = γm0c2

p= γm0u

E2 = pc( )2 + m0c

2( )2

β =vc≡pcE

NowlookatwhathappenstotheequaMonsintherightboxifm=0.

E = pcβ =1

NowlookatwhathappenstotheequaMonsinthelebboxifm=0.

E = γm0c2 ≈ ∞•0

p= γm0u

≈ ∞•0

Undefined

InRela9vis9cMechanicswecanhaveaparMclewithm=0.Itisaphoton.Chapter4

AphotonisaconsequenceofQuantumMechanics(Einstein).ItistheparMclenatureorlightwaves.Theenergycomesintermsofphotonseachwithadiscreteenergyandmomentumdependinguponthewavelengthofthelight.

Youmusthavedonetheexperimentwithasupportedplatewithonesidepolishedandtheothersideblack.Whenlightshinesonthistheplaterotatesbecauseofthemomentumchangefromthephotons.

Our example of ionization of Hγ + H → e+ p

PhotonsaremasslessparMclesv=c:E=pcNeutrinoswerebelievedtobemassless,butrecentexperimentsshowm~10‐5me.

Gravitonsarerelatedtogravitythewayphotonsaretolightshouldhavem=0.Noexperimentalevidence‐‐LIGO

Quiz

ConsidertheinelasMccollisionoftwoequalmassobjectsshowninthefigure.

NowconsidertheS’systemmovingwithobject#1atavelocityofv.

Provethatbyusingthenewdefini9onofmomentumthatmomentumisconservedintheS’coordinatesystem.

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