7.3 Solutions 413
V λ× 10 −^10 m(1/λ)× 106
1.48 3,660 2,732
1.15 4,050 2,469
0.93 4,360 2,293
0.62 4,920 2,032
0.36 5,460 1,831
0.24 5,790 1,727Slope= 1. 24 / 106 =hc/e
h= 1. 24 × 10 −^6 ×e/c=(1. 24 × 10 −^6 × 1. 6 × 10 −^19 /(3× 108 )= 6. 6 × 10 −^34 J-s
Intercept= 1. 5 × 10 −^6 m−^1
W=hc×intercept= 6. 6 × 10 −^34 × 3 × 108 × 1. 5 × 106
= 3 × 10 −^19 J= 3 × 10 −^19 / 1. 6 × 10 −^19 eV= 1 .9eV
hν 0 =W
ν 0 =W/h= 3 × 10 −^19 / 6. 6 × 10 −^34 = 4. 545 × 1014
λ 0 = 6. 6 × 10 −^7 m7.70 The solid angle subtended at the source
Ω=area of the face of the crystal/square of the distamce from the source
=πR^2 /d^2
Assuming that the gamma rays are emitted isotropically, the fraction of
gamma rays entering the crystal
F=Ω/ 4 π=R^2 / 4 d^2 = 22 / 4 × 1002 = 10 −^4
The number of photons entering the crystal per second from the source of
strengthSis
N=SF= 100 × 10 −^6 × 3. 7 × 1010 × 10 −^4
= 370 /s
Number of photons absorbed in the crystal of 1 cm thickness due to photo-
electric effect will be
Nph=N(1−exp(−μphx)=370(1−e−^0.^03 ×^1 )= 11
and the number absorbed due to Compton scattering will be
Nc=N(1−exp(−μcx)=370(1−e−^0.^24 ×^1 )= 79
Assuming that each photon that interacts in the crystal produces a photo
electron,NphandNcalso denote the number of photoelectrons in the respec-
tive processes.
Number of photons that do not interact in 1sec in the crystal is 370−(11+
79)= 280
7.71 In the first excited state the ionization energy is 13. 6 / 22 = 3 .4eV
T=hν−W
hν=T+W= 10. 7 + 3. 4 = 14 .1eV
In the ground state,W= 13 .6eV
T= 14. 1 − 13. 6 = 0 .5eV
7.72 T=hν−W(Einstein’s equation)
T 1 =hc/λ 1 −W