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Solution Manual for Modern Physics with Modern Computational Methods: for Scientists and Engineers 3rd Edition by John Morrison, ISBN: 9780128177907, All 15 Chapters Covered, Verified Latest Edition$16.49
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Solution Manual for Modern Physics with Modern Computational Methods: for Scientists and Engineers 3rd Edition by John Morrison, ISBN: 9780128177907, All 15 Chapters Covered, Verified Latest Edition
Solution Manual for Modern Physics with Modern Computational Methods: for Scientists and Engineers 3rd Edition by John Morrison, ISBN: 9780128177907, All 15 Chapters Covered, Verified Latest Edition Solution Manual for Modern Physics with Modern Computational Methods: for Scientists and Engineers 3...
1. TheFenergyFofFphotonsFinFtermsFofFtheFwavelengthFofFlightFi
sFgivenFbyFEq.F(1.5).FFollowingFExampleF 1.1FandFsubstitutingF
λF=F200FeVFgives:
hc 1240F eVF ·Fnm
= =F6.2FeV
EphotonF= λ 200Fnm
2. TheF energyF ofF theF beamF eachF secondF is:
power 100F W
= =F100FJ
EtotalF= time 1F s
TheFnumberFofFphotonsFcomesFfromFtheFtotalFenergyFdividedFb
yFtheFenergyFofFeachFphotonF(seeFProblemF1).FTheFphoton’sFene
rgyFmustFbeFconvertedFtoFJoulesFusingFtheFconstantF1.602F×F10
−19FJ/eVF,FseeFExampleF1.5.FTheFresultFis:
FE total F
N = = 100FJ =F1.01F×F1020
photons E
pho
ton 9.93F×F10−19
forF theF numberF ofF photonsF strikingF theF surfaceF eachF second.
3.WeFareFgivenFtheFpowerFofFtheFlaserFinFmilliwatts,FwhereF1Fm
WF=F10−3FWF.FTheFpowerFmayFbeFexpressedFas:F1FWF=F1FJ/s.F
FollowingFExampleF1.1,FtheFenergyFofFaFsingleFphotonFis:
1240F eVF ·Fnm
hcF =F1.960FeV
EphotonF = 632.8F nm
=
λF
F
er:
hc
=F(KE)maxF+FWF =F2.3F eVF +F0.9F eVF =F3.2F eV
λF
SolvingF Eq.F (1.5)F forF theF wavelength:
1240F eVF ·Fnm
λF= =F387.5Fnm
3.2F e
V
6. AFpotentialFenergyFofF0.72FeVFisFneededFtoFstopFtheFflowFofFelectro
ns.FHence,F(KE)maxFofFtheFphotoelectronsFcanFbeFnoFmoreFthanF0.
72FeV.FSolvingFEq.F(1.6)FforFtheFworkFfunction:
hc 1240F eVF ·Fn —F0.72F eVF =F1.98F eV
WF =F —
λ m
(KE)maxF
=
460Fnm
7. ReversingF theF procedureF fromF ProblemF 6,F weF startF withF Eq.F (1.6):
hcF 1240F eVF ·Fn
(KE)maxF = −FWF —F1.98F eVF =F3.19F eV
= m
λ
240Fnm
Hence,FaFstoppingFpotentialFofF3.19FeVFprohibitsFtheFelectronsFfro
mFreachingFtheFanode.
8. JustF atF threshold,F theF kineticF energyF ofF theF electronF isF z
ero.F SettingF(KE)maxF=F0F inF Eq.F (1.6),
hc
WF= = 1240F eVF ·Fn =F3.44F eV
λ0 m
360Fnm
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