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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