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
k k




1.kThekWave-ParticlekDuality

2.kThekSchrödingerkWavekEquation

3.kOperatorskandkWaves

4.kThekHydrogenkAtom

5.kMany-ElectronkAtoms

6.kThekEmergencekofkMaserskandkLasers

7.kDiatomickMolecules

8.kStatisticalkPhysics

9.kElectronickStructurekofkSolids

10.kChargekCarrierskinkSemiconductors

11.kSemiconductorkLasers

12.kThekSpecialkTheorykofkRelativity

13.kThekRelativistickWavekEquationskandkGeneralkRelativity

14.kParticlekPhysics

15.kNuclearkPhysics

,1

Thek Wave-Particlek Dualityk -k Solutions




1. Thekenergykofkphotonskinktermskofkthekwavelengthkofklightkisk
givenkbykEq.k(1.5).kFollowingkExamplek 1.1kandksubstitutingkλk
=k200keVkgives:
hc 1240k eVk ·knm
= =k6.2keV
Ephotonk= λ 200knm
2. Thek energyk ofk thek beamk eachk secondk is:
power 100k W
= =k100kJ
Etotalk= time 1k s
Theknumberkofkphotonskcomeskfromkthektotalkenergykdividedkby
kthekenergykofkeachkphotonk(seekProblemk1).kThekphoton’skenerg

ykmustkbekconvertedktokJouleskusingkthekconstantk1.602k×k10−19k
J/eVk,kseekExamplek1.5.kThekresultkis:
N =kEtotalk = 100kJ =k1.01k×k1020
photons E
pho
ton 9.93k×k10−19
fork thek numberk ofk photonsk strikingk thek surfacek eachk second.
3.Wekarekgivenkthekpowerkofktheklaserkinkmilliwatts,kwherek1kmW
=k10−3kWk.kThekpowerkmaykbekexpressedkas:k1kWk=k1kJ/s.kFollo
k

wingkExamplek1.1,kthekenergykofkaksinglekphotonkis:
1240k eVk ·knm
hck =k1.960keV
Ephotonk = 632.8k nm
=
λk
k


Wek nowk convertk tok SIk unitsk (seek Examplek 1.5):
1.960keVk×k1.602k×k10−19kJ/eVk =k3.14k×k10−19kJ
Followingkthek samekprocedurek ask Problemk2:
1k×k10−3kJ/s 15k photons
Ratekofk emissionk=k = k3.19k×k10
3.14k×k10−19k J/photonk s

, 4. Thekmaximumkkinetickenergykofkphotoelectronskiskfoundkusi
ngkEq.k(1.6)kandkthekworkkfunctions,kW,kofkthekmetalskarekgivenk
inkTablek1.1.kFollowingkProblemk 1,k Ephotonk=khc/λk=k6.20k eVk.k Fo
rk partk (a),k Nak hask Wk =k2.28k eVk:
(KE)maxk=k6.20keVk−k2.28keVk =k3.92keV
Similarly,kforkAlkmetalkinkpartk(b),kWk =k4.08keVk givingk(KE)maxk=k2.12keV
andkforkAgkmetalkinkpartk(c),kWk=k4.73keVk,kgivingk(KE)maxk=k1.47keVk.

5.Thiskproblemkagainkconcernskthekphotoelectrickeffect.kAskinkPro
blemk4,kwekusekEq.k(1.6):
hck−k
(KE)maxk =
Wkλ
wherek Wk isk thek workk functionk ofk thek materialk andk thek termk hc/λk d
escribeskthekenergykofkthekincomingkphotons.kSolvingkforktheklatter:
hc
=k(KE)maxk+kWk =k2.3k eVk +k0.9k eVk =k3.2k eV
λk
SolvingkEq.k(1.5)kforkthekwavelength:
1240k eVk ·knm
λk= =k387.5knm
3.2k e
V
6. Akpotentialkenergykofk0.72keVkiskneededktokstopkthekflowkofkelectron
s.kHence,k(KE)maxkofkthekphotoelectronskcankbeknokmorekthank0.72k
eV.kSolvingkEq.k(1.6)kforkthekworkkfunction:
hc 1240k eVk ·kn —k0.72k eVk =k1.98k eV
Wk =k —
λ m
(KE)maxk
=
460knm
7. Reversingk thek procedurek fromk Problemk 6,k wek startk withk Eq.k (1.6):
hck 1240k eVk ·kn
(KE)maxk = −kWk —k1.98k eVk =k3.19k eV
= m
λ
240knm
Hence,kakstoppingkpotentialkofk3.19keVkprohibitskthekelectronskfro
mkreachingkthekanode.

8. Justk atk threshold,k thek kinetick energyk ofk thek electronk isk zer
o.k Settingk(KE)maxk=k0k ink Eq.k (1.6),
hc
Wk= = 1240k eVk ·kn =k3.44k eV
λ0 m

360knm
9. Akfrequencykofk1200kTHzkiskequalktok1200k×k1012kHz.kUsingkEq.k(1.10),