1000 Solved Problems in Modern Physics

5.3 Solutions 307

(b) p(E)=

1

e−^1.^932 + 1

= 0. 873

(c) p(E)=

1

e^0 + 1

= 0. 5

5.28 Assuming that the Fermi energy is to be at the middle of the gap between the
conduction and valence bands,E−EF=^1 / 2 Eg
p(E)=

1

e(E−EF)/kT+^1

=

1

eEg/^2 kT+ 1

The factorEg/ 2 kT=

1. 1

2 × 8. 625 × 10 −^5 × 400

= 15. 942

p(E)≈e−^15.^942 = 8. 4 × 10 −^6

5.29 The Debye temperatureθis

θ=

hνm

k

νm=

k

h

θ=

1. 38 × 10 −^23 × 360

6. 625 × 10 −^34

= 7. 5 × 1012 Hz

5.30 (a) At high temperaturesT>> θE, in the denominator (eθE/T−1)^2 ≈θE^2 /T^2 ,

and in the numeratoreθE/T→1, so thatCv→ 3 N 0 k= 3 R, the Dulong –

Petit’s value

(b) When the temperature is very lowT << θE, and in the bracket of the

denominator, 1 is negligible in comparison with the exponential term.

Therefore,Cv → 3 R(θE/T)^2 e−θE/T. Thus the specific heat goes to zero

asT→0. However, the experimentally observed specific heats at low

temperatures decrease more gradually than the exponential decrease sug-

gested by Einstein’s formula.

5.31Cv=

9 R

x^3

∫x

0

ξ^4 eξ

(eξ−1)^2

dξ (1)

This equation may be integrated by parts,

∫x

0

ξ^4 eξ

(eξ−1)^2

dξ=−

∫

ξ^4

d

dξ

(

1

eξ− 1

)

dξ

=−ξ^4

1

eξ− 1

+

∫

1

eξ− 1

dξ^4

dξ

dξ

=−ξ^4

1

eξ− 1

+ 4

∫

ξ^3

eξ− 1

dξ

Thus (1) becomes

Cv= 9 R

[

4

x^3

∫x

0

ξ^3

eξ− 1

dξ−

x

ex− 1

]

(2)