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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 $15.49   Add to cart
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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
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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 3...

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

,Table of contents
U U




1.UTheUWave-ParticleUDuality

2.UTheUSchrödingerUWaveUEquation

3.UOperatorsUandUWaves

4.UTheUHydrogenUAtom

5.UMany-ElectronUAtoms

6.UTheUEmergenceUofUMasersUandULasers

7.UDiatomicUMolecules

8.UStatisticalUPhysics

9.UElectronicUStructureUofUSolids

10.UChargeUCarriersUinUSemiconductors

11.USemiconductorULasers

12.UTheUSpecialUTheoryUofURelativity

13.UTheURelativisticUWaveUEquationsUandUGeneralURelativity

14.UParticleUPhysics

15.UNuclearUPhysics

,1

TheU Wave-ParticleU DualityU -U Solutions




1. TheUenergyUofUphotonsUinUtermsUofUtheUwavelengthUofUlight
UisUgivenUbyUEq.U(1.5).UFollowingUExampleU 1.1UandUsubstituti
ngUλU=U200UeVUgives:
hc 1240U eVU ·Unm
= =U6.2UeV
EphotonU= λ 200Unm
2. TheU energyU ofU theU beamU eachU secondU is:
power 100U W
= =U100UJ
EtotalU= time 1U s
TheUnumberUofUphotonsUcomesUfromUtheUtotalUenergyUdividedU
byUtheUenergyUofUeachUphotonU(seeUProblemU1).UTheUphoton’sU
energyUmustUbeUconvertedUtoUJoulesUusingUtheUconstantU1.602
U×U10
−19UJ/eVU,UseeUExampleU1.5.UTheUresultUis:

UE total U
N = = 100UJ =U1.01U×U1020
photons E
pho
ton 9.93U×U10−19
forU theU numberU ofU photonsU strikingU theU surfaceU eachU second.
3.WeUareUgivenUtheUpowerUofUtheUlaserUinUmilliwatts,UwhereU1U
mWU=U10−3UWU.UTheUpowerUmayUbeUexpressedUas:U1UWU=U1UJ
/s.UFollowingUExampleU1.1,UtheUenergyUofUaUsingleUphotonUis
:
1240U eVU ·Unm
hcU =U1.960UeV
EphotonU = 632.8U nm
=
λU
U



WeU nowU convertU toU SIU unitsU (seeU ExampleU 1.5):
1.960UeVU×U1.602U×U10−19UJ/eVU =U3.14U×U10−19UJ
FollowingU theU sameU procedureU asU ProblemU 2:
1U×U10−3UJ/s 15U photons
RateUofU emissionU=U = U3.19U× U10
3.14U×U10−19U J/photonU s

, 2

4. TheUmaximumUkineticUenergyUofUphotoelectronsUisUfoundUu
singUEq.U(1.6)UandUtheUworkUfunctions,UW,UofUtheUmetalsUareUg
ivenUinUTableU1.1.UFollowingUProblemU 1,U EphotonU=Uhc/λU=U6.20
U eV U.U ForU partU (a),U NaU hasU W U =U2.28U eV U:



(KE)maxU=U6.20UeVU−U2.28UeVU =U3.92UeV
Similarly,UforUAlUmetalUinUpartU(b),UWU =U4.08UeVU givingU(KE)maxU=U2.12
U eV

andUforUAgUmetalUinUpartU(c),UWU=U4.73UeVU,UgivingU(KE)maxU=U1.47UeVU.

5.ThisUproblemUagainUconcernsUtheUphotoelectricUeffect.UAsUinUP
roblemU4,UweUuseUEq.U(1.6):
hcU−U
(KE)maxU =
WUλ
whereU WU isU theU workU functionU ofU theU materialU andU theU termU hc/
λU describesUtheUenergyUofUtheUincomingUphotons.USolvingUforUtheUl
atter:
hc
=U(KE)maxU+UWU =U2.3UeVU +U0.9U eVU =U3.2UeV
λU
SolvingU Eq.U (1.5)U forU theU wavelength:
1240U eVU ·Unm
λU= =U387.5Unm
3.2U e
V
6. AUpotentialUenergyUofU0.72UeVUisUneededUtoUstopUtheUflowUofUelect
rons.UHence,U(KE)maxUofUtheUphotoelectronsUcanUbeUnoUmoreUtha
nU0.72UeV.USolvingUEq.U(1.6)UforUtheUworkUfunction:
hc 1240U eVU ·U —U0.72U eVU =U1.98U eV
WU =U —
λ nm
(KE)max
U =

460Unm
7. ReversingU theU procedureU fromU ProblemU 6,U weU startU withU Eq.U (1.6):
hcU 1240U eVU ·U
(KE)maxU = −UWU —U1.98U eVU =U3.19U eV
= nm
λ
240Unm
Hence,UaUstoppingUpotentialUofU3.19UeVUprohibitsUtheUelectronsUfr
omUreachingUtheUanode.

8. JustU atU threshold,U theU kineticU energyU ofU theU electronU isU
zero.U SettingU(KE)maxU=U0U inU Eq.U (1.6),
hc
WU= = 1240U eVU ·U =U3.44U eV
λ0 nm

360Unm

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