SOLUTION MANUAL Modern Physics with Modern Computational Methods: for Scientists and Engineers 3rd Edition by Morrison Chapters 1- 15

SOLUTION MANUAL Modern Physics with Modern
Computational Methods: for Scientists and Engineers
3rd Edition by Morrison Chapters 1- 15

,Table of contents
1.RTheRWave-ParticleRDuality

2.RTheRSchrödingerRWaveREquation

3.ROperatorsRandRWaves

4.RTheRHydrogenRAtom

5.RMany-ElectronRAtoms

6.RTheREmergenceRofRMasersRandRLasers

7.RDiatomicRMolecules

8.RStatisticalRPhysics

9.RElectronicRStructureRofRSolids

10.RChargeRCarriersRinRSemiconductors

11.RSemiconductorRLasers

12.RTheRSpecialRTheoryRofRRelativity

13.RTheRRelativisticRWaveREquationsRandRGeneralRRelativity

14.RParticleRPhysics

15.RNuclearRPhysics

,1

TheR Wave-ParticleR DualityR -R Solutions




1. TheRenergyRofRphotonsRinRtermsRofRtheRwavelengthRofRlightRi
sRgivenRbyREq.R(1.5).RFollowingRExampleR 1.1RandRsubstituting
RλR=R200ReVRgives:


hc 1240R eVR ·Rnm
= =R6.2ReV
EphotonR= λ 200Rnm
2. TheR energyR ofR theR beamR eachR secondR is:
power 100R W
= =R100RJ
EtotalR= time 1R s
TheRnumberRofRphotonsRcomesRfromRtheRtotalRenergyRdividedR
byRtheRenergyRofReachRphotonR(seeRProblemR1).RTheRphoton’sRe
nergyRmustRbeRconvertedRtoRJoulesRusingRtheRconstantR1.602R×
R10
−19 RJ/eVR,RseeRExampleR1.5.RTheRresultRis:

REtotal R
N = = 100RJ =R1.01R×R1020
photons E
pho
ton 9.93R×R10−19
forR theR numberR ofR photonsR strikingR theR surfaceR eachR second.
3.WeRareRgivenRtheRpowerRofRtheRlaserRinRmilliwatts,RwhereR1Rm
WR=R10−3RWR.RTheRpowerRmayRbeRexpressedRas:R1RWR=R1RJ/s.R
FollowingRExampleR1.1,RtheRenergyRofRaRsingleRphotonRis:
1240R eVR ·Rnm
hcR =R1.960ReV
EphotonR = 632.8R nm
=
λR
R


WeR nowR convertR toR SIR unitsR (seeR ExampleR 1.5):
1.960ReVR×R1.602R×R10−19RJ/eVR =R3.14R×R10−19RJ
FollowingRtheRsameR procedureRasRProblemR 2:
1R×R10−3RJ/s 15R photons
RateR ofR emissionR=R = R3.19R× R10
3.14R×R10−19R J/photonR s

, 2

4. TheRmaximumRkineticRenergyRofRphotoelectronsRisRfoundRus
ingREq.R(1.6)RandRtheRworkRfunctions,RW,RofRtheRmetalsRareRgiv
enRinRTableR1.1.RFollowingRProblemR 1,R EphotonR=Rhc/λR=R6.20R e
VR.R ForR partR (a),R NaR hasR WR =R2.28R eVR:
(KE)maxR=R6.20ReVR−R2.28ReVR =R3.92ReV
Similarly,RforRAlR metalRinRpartR (b),R WR =R4.08ReVR givingR(KE)maxR=R2.12Re
V
andRforRAgRmetalRinRpartR(c),RWR=R4.73ReVR,RgivingR(KE)maxR=R1.47ReVR.

5.ThisRproblemRagainRconcernsRtheRphotoelectricReffect.RAsRinRPr
oblemR4,RweRuseREq.R(1.6):
hcR−R
(KE)maxR =
WRλ
whereR WR isR theR workR functionR ofR theR materialR andR theR termR hc/
λR describesRtheRenergyRofRtheRincomingRphotons.RSolvingRforRtheRla
tter:
hc
=R(KE)maxR+RWR =R2.3R eVR +R0.9R eVR =R3.2R eV
λR
SolvingREq.R(1.5)RforR theRwavelength:
1240R eVR ·Rnm
λR= =R387.5Rnm
3.2R e
V
6. ARpotentialRenergyRofR0.72ReVRisRneededRtoRstopRtheRflowRofRelectr
ons.RHence,R(KE)maxRofRtheRphotoelectronsRcanRbeRnoRmoreRthanR
0.72ReV.RSolvingREq.R(1.6)RforRtheRworkRfunction:
hc 1240R eVR ·Rn —R0.72R eVR =R1.98R eV
WR =R —
λ m
(KE)maxR
=
460Rnm
7. ReversingR theR procedureR fromR ProblemR 6,R weR startR withR Eq.R (1.6):
hcR 1240R eVR ·Rn
(KE)maxR = −RWR —R1.98R eVR =R3.19R eV
= m
λ
240Rnm
Hence,RaRstoppingRpotentialRofR3.19ReVRprohibitsRtheRelectronsRfr
omRreachingRtheRanode.

8. JustR atR threshold,R theR kineticR energyR ofR theR electronR isR z
ero.R SettingR(KE)maxR=R0R inR Eq.R (1.6),
hc
WR= = 1240R eVR ·Rn =R3.44R eV
λ0 m

360Rnm