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PHYSICS201W03EXAM3SOLUTIONS

Problem1(SpecialTheoryofRelativity)

AnadvancedspacecrafttravelspastEarthandMarsinastraightlineat speedv=0:8catatimewhenEarthandMarsare2:41011mapart. Thedistanceismeasuredinthexedframeofreference,inwhichthe Sun,Earth,andMarsareatrest.(Weneglectanymotionoftheplanets withrespecttoeachotherandtheSunduringthisshorttime.) (A)Inthexedreferenceframe,howlongdoesittakeforthespace- crafttogofromEarthtoMars? ThemeasurementofthedistancebetweenEarthandMarsistakenbya"ruler" atrestintheEarthandMarsframe,andsoistheproperlength:Lp=2:41011m. Thusifthevelocityoftheshipisv=0:8cintheEarth-Marsframe,theshipwill take t=Lp v=2:41011s0:8(3108m=s)=1000s (B)Intheframeinwhichthespacecraftisatrest,howfastarethe

EarthandMarsmoving?

TheshipwillseetheEarthandMarsmovingatv=0:8c.

(C)Inthespacecraftframe,howfarapartaretheEarthandMars? Sinceweknowthatthemeasurementgiven,2:41011mistheproperlength (the"ruler"bywhichthiswasmeasuredisatrestintheEarth-Marsframe),then weknowthattheshipwillseethedistancebetweenEarthandMarscontracted: L=Lp =2:41011m5=3=35(2:41011m)=1:441011m Where =1 q

1v2c2=1

q

1(0:8c)2c2=53

(D)Inthespacecraftframe,howmuchtimeelapsesbetweentheEarth passingby,andMarspassingby? Wecansolvethistwoways.Weknowthatthepropertimemeasurementistaken onthespaceshipbecausethetwoeventsbywhichwemeasurethetimeinterval, thetimethespaceshipleavesEarthandthetimetheshipreachesmars,appear tohappenatthesameplaceintheshipframe{thatis,itlookstoanobserveron theshiplikeEarthisgoingawayfromthemandmarsisgoingtowardsthem.Thus theymeasurepropertimeandwecanusethetimedilationequation: tp=t =351000s=600s

WecanalsousethesamemethodasinpartA:

t0=L v=1:4410110:8(3108m=s)=600s 1

2PHYSICS201W03EXAM3SOLUTIONS

Problem2(RelativisticMomentumandEnergy)

Electronsandprotonsareacceleratedthroughavoltagedierence of109V.Therestenergiesoftheelectronandprotonare0:511MeV and938MeV,respectively. (A)Calculatetherelativisticmomentaofelectronsandprotons Therestenergiesaremec2=0:511MeVfortheelectron,andmpc2=938MeV fortheproton. Whenachargedparticleisacceleratedthroughavoltagedierenceof109V,its kineticenergywillbeQ(109V)whereQisthechargeoftheparticle.Inthiscase, boththeprotonandtheelectronhaveachargee(theprotonis+eandtheelectron ise)sobothparticleswillendupwithkineticenergy109eV(rememberinthe photoelectriceecthowwerelatedthekineticenergywiththestoppingpotential{ thisisthesameidea).WecanthususetheequationE2=c2p2+(mc2)2: E e=Ke+mec2E2e=c2p2 e+(mec2)2 E p=Kp+mpc2E2p=c2p2 p+(mpc2)2 Thus cp e=p

E2e(mec2)2=pK2e+2mec2Ke=Kes1+2mec2Ke

=109eVr

1+2(0:511106eV)109eV=1:0005109eV

Fortheprotonwegetthesameresult:

cp p=Kps

1+2mpc2Kp=109eVr1+2(938106eV)109=1:696109eV

Alternativesolution:

K e=109eV=(

1)mec2=(

1)0:511106eV=)

=1957:95=1 q

1v2c2=)v=0:99999987c

Thuscpe=c(

mv)=1957:95(0:99999987)mec2=1:0005109eV

Similarlyfortheproton,

K p=109eV=(

1)mpc2=(

1)938106eV=)

=2:066=1 q

1v2c2=)v=0:875c

Thuscpp=c(

mv)=2:066(0:875)mpc2=1:696109eV (B)Calculatethetotalenergyoftheseelectronsandprotons. E e=Ke+mec2=1000:511106eV E p=Kp+mpc2=1938106eV

PHYSICS201W03EXAM3SOLUTIONS3

Problem3(ComptonEffect)

InaComptonscatteringevent,thescatteredphotonhasanenergy of150keVandtherecoilingelectronhasakineticenergyof40keV.

Assumethatbeforethescatteringtheelectronisatrest.

(A)Findthewavelength(innanometers)andtheenergy(inkeV)of theincidentphoton. Energyconservationtellsusthatthetotalenergyofthissystembeforethe electronishitbythephotonisthesameasthetotalenergyofthesystemafterthe photonhitstheelectron. Ignoringtherestenergy,becauseitdoesn'tchange,wehave E i=Ef hc i+0=hcf+Ke hc i=150103eV+40103eV

1240eV(nm)

i=190103eV

1240eV(nm)

190103eV=i

0:0065nm=i

(B)Firstndthewavelength(innanometers)ofthescatteredphoton andthencalculatethescatteringangle. hc f=150103eV=)f=hc150103eV=1240eV(nm)150103eV=0:0083nm

NowwecanapplytheComptonshiftequation:

=fi=0:0083nm0:0065nm=0:0018=c(1cos)

Wesolveforcos:

cos=1fi c=10:0018nm0:0024nm=0:25

Taketheinverse-cosinetond=72:5.

4PHYSICS201W03EXAM3SOLUTIONS

Problem4(DeBroglieWavelength)

TheelectronsinDavisson-Germerexperimenthadakineticenergy of54eV. (A)CalculatetheDeBrogliewavelengthoftheseelectrons. Thekineticenergyoftheelectronsismuchlessthantherestenergyofthe electron,54eV0:5116eV,sowecanusetheclassicalformulawhichrelateskinetic energy(K=mv2=2i)tomomentum(p=mv): K=1

2mv2=12mm2v2=p22m=)p=p2mK

UsingtheDeBrogliewaveequation,wehave

e=h p hc cp2meK hc p2mec2K

1240eV(nm)

p2(0:511106eV)54eV

1240nm

7428:86

=0:167nm (B)WhatwouldbetheDeBrogliewavelengthofprotonsofthesame kineticenergyof54eV? Again,sincethekineticenergyismuchlessthantherestenergyofprotons,we canusetheclassicalformulaforkineticenergyand,asinpartA: p=hc p2mpc2K=1240eV(nm)p2(938106)54eV=0:003896nm Andsowenotethatthemoremassiveobjectshavemuchsmallerwavelengths thanlessmassiveobjectswiththesamekineticenergy.quotesdbs_dbs14.pdfusesText_20