1000 Solved Problems in Modern Physics

308 5 Solid State Physics

(a) At high temperatures,θD>>T,orx<<1, and the exponential can be

expanded to give

Cv= 9 R

(

4

3

− 1

)

= 3 R (Dulong Petit’s law)

(b) At very low temperaturesT<< θDx>>1, (2) can be approximated to

Cv=9R

4

x^3

∫∞

0

ξ^3 dξ

eξ− 1

=

12

5

π^4

(

T

θD

) 3

where the value of the integral isπ^4 /15. Thus,Cv∝T^3

5.32 If there areNfree electrons in the metal there will beN/2 occupied quantum
states at the absolute zero of temperature in accordance with the Fermi Dirac
statistics. In Fermi-Dirac statistics at absolute zero, kinetic energy is not zero
as would be required if the Boltzmann statistics were assumed.
AsN(E)dEgives the number of states per unit volume, in a crystal of volume
V, the number of electrons in the range fromEtoE+dEis

2 V·

2 π

h^3

(2m)^3 /^2 E^1 /^2 dE (1)

The total energy of these electrons would be

Etotal=

∫Emax

0

4 πV

h^3

(2m)^3 /^2 E^3 /^2 dE=

4 πV(2m)

(^32)
h^3

·

2

5

Emax^5 /^2 (2)

But,

Emax=

h^2

8 m

(

3 N

πV

) 2 / 3

(3)

Combining (2) and (3),

Etotal=

3

5

NEmax (4)

or per electron 3Emax/5. The quantityEmax=EF, the Fermi energy

5.33 The density of statesn(E) (the number of states per unit volume of the solid
in the unit energy interval) is given by

n(E)=

8

√

2 πm^3 /^2

h^3

E^1 /^2

=

(8

√

2 π)(9. 11 × 10 −^31 )^3 /^2

(6. 63 × 10 −^34 )^3

(4× 1. 6 × 10 −^19 )^1 /^2

= 8. 478 × 1046 m−^3 J−^1 = 1. 356 × 1028 m−^3 eV−^1

Number of statesNthat lie in the rangeE = 4 .00eV toE= 4 .01eV, for

volume,V=a^3