Modern Physics Questions & Answers

f f f f f




f




b
f b f f f f f

, f f




1. The bWave-Particle bDuality

2. The bSchrödinger bWave bEquation

3. Operators band bWaves

4. The bHydrogen bAtom

5. Many-Electron bAtoms

6. The bEmergence bof bMasers band bLasers

7. Diatomic bMolecules

8. Statistical bPhysics

9. Electronic bStructure bof bSolids

10. Charge bCarriers bin bSemiconductors

11. Semiconductor bLasers

12. The bSpecial bTheory bof bRelativity

13. The bRelativistic bWave bEquations band bGeneral bRelativity

14. Particle bPhysics

15. Nuclear bPhysics

, 1

The bWave-Particle bDuality b- bSolutions




1. The benergy bofphotons bin bterms bofthe bwavelength bof
blightis bgiven bby bEq. b(1.5). bFollowing bExample b 1.1
band bsubstituting bλ b= b200 beV bgives:
hc b 1240 beV b · bnm
= =6.2eV
Ephoton b= b λ 200 bnm
2.The benergy b of b the b beam beach b second b is:
power 100 bW
= =100 bJ
Etotal b= time 1
bs
The bnumber bof bphotons bcomes bfrom bthe btotal benergy
bdivided bby bthe benergy bof beach bphoton b(see bProblem
b1). bThe bphoton’s benergy bmust bbe bconverted bto
bJoules busing bthe bconstant b1.602 b× b10−19 bJ/eV, bsee
bExample b1.5. bThe bresult bis:
N =Etotal = 100J =1.01×1020
photons b
Epho b
bton 9.93×10−19
for b the b number b of b photons b striking b the b surface b each b second.
3.Wearegiven bthepowerof bthelaser bin bmilliwatts, bwhere
b1mW b = b10−3 bW. bThe bpower bmay bbe bexpressed bas: b1
bW b= b1 bJ/s. bFollowing bExample b1.1, bthe benergy bof ba
bsingle bphoton bis:
1240 beV b · bnm
hc =1.960 beV
Ephoton 632.8
b= bnm
=
λ b
We bnow bconvert bto bSI bunits b(see bExample b1.5):
1.960eV b×1.602×10−19 bJ/eV b =3.14×10−19 b J
Following bthe bsame bprocedure bas bProblem b2:
1×10−3 bJ/s 15
b photons bRate bof b emission b= b 3.14×10−19
b J/photon b = b3.19×10 bs

, 2

4. The bmaximum bkinetic benergy bof bphotoelectrons bis
bfound busing bEq. b(1.6) band bthe bwork bfunctions, bW,
bof bthe bmetals bare bgiven bin bTable b1.1. bFollowing
bProblem b 1, b Ephoton b = bhc/λ b= b6.20 beV b. b For b part
b (a), b Na b has b W b = b2.28 b eV b:
(KE)max=6.20eV b−2.28 beV b =3.92 beV
Similarly, bfor bAl bmetal bin bpart b(b), bW b= b4.08 beV b giving b(KE)max b = b2.12
beV
andfor bAg bmetal bin bpart b(c), bW b=4.73 beV, bgiving b(KE)max b =1.47 beV.
5.This bproblem bagain bconcerns bthe bphotoelectric beffect.
bAs bin bProblem b4, bwe buse bEq. b(1.6):
hc
(KE)max = b −
Wλ
where b W b is b the b work b function b of b the b material b and b the b term b hc/λ
describes bthe benergy bof bthe bincomingphotons. bSolvingfor bthe blatter:
hc
λ b (KE)max + W b= b2.3 beV b +0.9 beV b = b3.2 beV
= f




Solving b Eq. b (1.5) b for b the b wavelength:
1240 beV b · bnm
λ= =387.5 bnm
3.2
eV
6. A bpotential benergy bof b0.72 beV bis bneeded bto bstop bthe bflow
bof belectrons. bHence, b (KE)max b of bthe bphotoelectrons bcan
bbenomore bthan b0.72 beV. bSolvingEq.(1.6) bfor bthe bwork
bfunction:
hc 1240 beV — b0.72 beV b= b1.98 beV
bW b= (KE) max
bλ b·
= nm

460 bnm
7. Reversing bthe bprocedure bfrom bProblem b6, bwe bstart bwith bEq. b(1.6):
hc b− 1240 beV
(KE)max — b1.98 beV b= b3.19 beV
b= bW b·
=
fb nm
λ
240 bnm
Hence, ba bstopping bpotential bof b3.19 beV bprohibits bthe
belectrons bfrom breaching bthe banode.

8. Just b at b threshold, b the b kinetic b energy b of b the
b electron b is bzero. b Setting b(KE)max b= b0 b in b Eq.
b (1.6),
hc b λ0
W= b = 1240 beV b ·