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
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
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