Solution Manual - Modern Physics with Modern Computational Methods: for Scientists and Engineers 3rd Edition by John Morrison, 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
D D




1.DTheDWave-ParticleDDuality

2.DTheDSchrödingerDWaveDEquation

3.DOperatorsDandDWaves

4.DTheDHydrogenDAtom

5.DMany-ElectronDAtoms

6.DTheDEmergenceDofDMasersDandDLasers

7.DDiatomicDMolecules

8.DStatisticalDPhysics

9.DElectronicDStructureDofDSolids

10.DChargeDCarriersDinDSemiconductors

11.DSemiconductorDLasers

12.DTheDSpecialDTheoryDofDRelativity

13.DTheDRelativisticDWaveDEquationsDandDGeneralDRelativity

14.DParticleDPhysics

15.DNuclearDPhysics

,1

TheD Wave-ParticleD DualityD -D Solutions




1. TheDenergyDofDphotonsDinDtermsDofDtheDwavelengthDofDlightDisDg
ivenDbyDEq.D(1.5).DFollowingDExampleD 1.1DandDsubstitutingDλD=D
200DeVDgives:
hc 1240D eVD ·Dnm
= =D6.2DeV
EphotonD= λ 200Dnm
2. TheD energyD ofD theD beamD eachD secondD is:
power 100D W
= =D100DJ
EtotalD= time 1D s
TheDnumberDofDphotonsDcomesDfromDtheDtotalDenergyDdividedDbyD
theDenergyDofDeachDphotonD(seeDProblemD1).DTheDphoton’sDenergyD
mustDbeDconvertedDtoDJoulesDusingDtheDconstantD1.602D×D10−19DJ/
eVD,DseeDExampleD1.5.DTheDresultDis:
N =DEtotalD = 100DJ =D1.01D×D1020
photons E
pho
ton 9.93D×D10−19
forD theD numberD ofD photonsD strikingD theD surfaceD eachD second.
3.WeDareDgivenDtheDpowerDofDtheDlaserDinDmilliwatts,DwhereD1DmWD
=D10−3DWD.DTheDpowerDmayDbeDexpressedDas:D1DWD=D1DJ/s.DFollow
ingDExampleD1.1,DtheDenergyDofDaDsingleDphotonDis:
1240D eVD ·Dnm
hcD =D1.960DeV
EphotonD =D 632.8D nm
=
λD
WeD nowD convertD toD SID unitsD (seeD ExampleD 1.5):
1.960DeVD×D1.602D×D10−19DJ/eVD =D3.14D×D10−19DJ
FollowingDtheDsameDprocedureDasDProblemD2:
1D×D10−3DJ/s 15D photons
RateDofD emissionD=D = D3.19D×D10
3.14D×D10−19D J/photonD s

, 4.TheDmaximumDkineticDenergyDofDphotoelectronsDisDfoundDusin
gDEq.D(1.6)DandDtheDworkDfunctions,DW,DofDtheDmetalsDareDgivenDin
DTableD1.1.DFollowingDProblemD 1,D Ephoton D=Dhc/λD=D6.20D eVD.D ForD

partD (a),D NaD hasD WD =D2.28D eVD:
(KE)maxD=D6.20DeVD−D2.28DeVD =D3.92DeV
Similarly,DforDAlDmetalDinDpartD(b),DWD =D4.08DeVD givingD(KE)maxD=D2.12DeV
andDforDAgDmetalDinDpartD(c),DWD =D4.73DeVD,DgivingD(KE)maxD=D1.47DeVD.

5.ThisDproblemDagainDconcernsDtheDphotoelectricDeffect.DAsDinDProb
lemD4,DweDuseDEq.D(1.6):
hcD−D
(KE)maxD =
WDλ
whereD WD isD theD workD functionD ofD theD materialD andD theD termD hc/λD d
escribesDtheDenergyDofDtheDincomingDphotons.DSolvingDforDtheDlatter:
hc
=D(KE)maxD+DWD =D2.3D eVD +D0.9DeVD =D3.2DeV
λD
SolvingD Eq.D (1.5)D forD theD wavelength:
1240D eVD ·Dnm
λD= =D387.5Dnm
3.2D e
V
6. ADpotentialDenergyDofD0.72DeVDisDneededDtoDstopDtheDflowDofDelectrons.
DHence,D(KE)max DofDtheDphotoelectronsDcanDbeDnoDmoreDthanD0.72DeV

.DSolvingDEq.D(1.6)DforDtheDworkDfunction:
hc 1240D eVD ·Dn —D0.72D eVD =D1.98D eV
W D =D —
λ m
(KE)maxD
=
460Dnm
7. ReversingD theD procedureD fromD ProblemD 6,D weD startD withD Eq.D (1.6):
hcD 1240D eVD ·Dn
(KE)maxD = −DWD —D1.98D eVD =D3.19D eV
= m
λ
240Dnm
Hence,DaDstoppingDpotentialDofD3.19DeVDprohibitsDtheDelectronsDfrom
DreachingDtheDanode.



8. JustD atD threshold,D theD kineticD energyD ofD theD electronD isD zer
o.D SettingD(KE)maxD=D0D inD Eq.D (1.6),
hc
WD= = 1240D eVD ·Dn =D3.44D eV
λ0 m

360Dnm
9. ADfrequencyDofD1200DTHzDisDequalDtoD1200D×D1012D Hz.DUsingDEq.D(1.10),