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

,Table of contents
p p



1.pThepWave-ParticlepDuality

2.pThepSchrödingerpWavepEquation

3.pOperatorspandpWaves

4.pThepHydrogenpAtom

5.pMany-ElectronpAtoms

6.pThepEmergencepofpMaserspandpLasers

7.pDiatomicpMolecules

8.pStatisticalpPhysics

9.pElectronicpStructurepofpSolids

10.pChargepCarrierspinpSemiconductors

11.pSemiconductorpLasers

12.pThepSpecialpTheorypofpRelativity

13.pThepRelativisticpWavepEquationspandpGeneralpRelativity

14.pParticlepPhysics

15.pNuclearpPhysics

,1

Thep Wave-Particlep Dualityp -p Solutions




1. Thepenergypofpphotonspinptermspofpthepwavelengthpofplightpispg
ivenpbypEq.p(1.5).pFollowingpExamplep 1.1pandpsubstitutingpλp=p
200peVpgives:
hc 1240p eVp ·pnm
= =p6.2peV
Ephotonp= λ 200pnm
2. Thep energyp ofp thep beamp eachp secondp is:
power 100p W
= =p100pJ
Etotalp= time 1p s
Thepnumberpofpphotonspcomespfromptheptotalpenergypdividedpbyp
thepenergypofpeachpphotonp(seepProblemp1).pThepphoton’spenergy
pmustpbepconvertedptopJoulespusingpthepconstantp1.602p×p10−19pJ

/eVp,pseepExamplep1.5.pThepresultpis:
N =pEtotalp = 100pJ =p1.01p×p1020
photons E
pho
ton 9.93p×p10−19
forp thep numberp ofp photonsp strikingp thep surfacep eachp second.
3.Weparepgivenptheppowerpofptheplaserpinpmilliwatts,pwherep1pmWp
=p10−3pWp.pTheppowerpmaypbepexpressedpas:p1pWp=p1pJ/s.pFollo
wingpExamplep1.1,pthepenergypofpapsinglepphotonpis:
1240p eVp ·pnm
hcp =p1.960peV
Ephotonp = 632.8p nm
p =
λp
Wep nowp convertp top SIp unitsp (seep Examplep 1.5):
1.960peVp×p1.602p×p10−19pJ/eVp =p3.14p×p10−19pJ
Followingp thep samep procedurep aspProblemp 2:
1p×p10−3pJ/s 15p photons
Ratep ofp emissionp=p = p3.19p×p10
3.14p×p10−19p J/photonp s

, 2

4.Thepmaximumpkineticpenergypofpphotoelectronspispfoundpusin
gpEq.p(1.6)pandpthepworkpfunctions,pW,pofpthepmetalsparepgivenpin
pTablep1.1.pFollowingpProblemp 1,p Ephotonp=phc/λp=p6.20p eV p.p Forp

partp (a),p Nap hasp Wp =p2.28p eVp:
(KE)maxp=p6.20peVp−p2.28peVp =p3.92peV
Similarly,p forp Alp metalp inp partp (b),p Wp =p4.08p eVp givingp(KE)maxp=p2.12p eV
andpforpAgpmetalpinppartp(c),pWp =p4.73peVp,pgivingp(KE)maxp=p1.47peVp.

5.Thispproblempagainpconcernspthepphotoelectricpeffect.pAspinpPro
blemp4,pwepusepEq.p(1.6):
hcp−p
(KE)maxp =
Wpλ
wherep Wp isp thep workp functionp ofp thep materialp andp thep termp hc/λp d
escribespthepenergypofpthepincomingpphotons.pSolvingpforptheplatter:
hc
=p(KE)maxp+pWp =p2.3peVp +p0.9peVp =p3.2peV
λp
Solvingp Eq.p (1.5)p forp thep wavelength:
1240p eVp ·pnm
λp= =p387.5pnm
3.2p e
V
6. Appotentialpenergypofp0.72peVpispneededptopstoppthepflowpofpelectrons
.pHence,p(KE)maxpofpthepphotoelectronspcanpbepnopmorepthanp0.72pe
V.pSolvingpEq.p(1.6)pforpthepworkpfunction:
hc 1240p eVp ·pn —p0.72p eVp =p1.98p eV
W p =p —
λ m
(KE)maxp
=
460pnm
7. Reversingp thep procedurep fromp Problemp 6,p wep startp withp Eq.p (1.6):
hcp 1240p eVp ·pn
(KE)maxp = −pWp —p1.98p eVp =p3.19p eV
= m
λ
240pnm
Hence,papstoppingppotentialpofp3.19peVpprohibitspthepelectronspfrom
preachingpthepanode.



8. Justp atp threshold,p thep kineticp energyp ofp thep electronp isp zer
o.p Settingp(KE)maxp=p0p inp Eq.p (1.6),
hc
Wp= = 1240p eVp ·pn =p3.44p eV
λ0 m

360pnm
9. Apfrequencypofp1200pTHzpispequalptop1200p×p1012p Hz.pUsingpEq.p(1.10),