Problems and Solutions on Thermodynamics and Statistical Mechanics

Statistid Physic8 277

where kTz = -p + E. As p/kT >> 1, we can substitute 00 for the upper
limit of the second integral in above expression so that

(a) Let f(e) = c312, then the internal energy E - I, C, = - -

TI i.e., n = 1. In fact, when T = 0 K, because the heat energy is so small,
only those electrons which lie in the transition band of width about kT on
the Fermi surface can be excited into energy levels of energies = kT. The
part of the internal energy directly related to T is then

(::IV

• NT - T2, i.e., C, - T.

(b) Let f(~) = d12, then M - I, hence M = Mo(1 -aT2), i.e., n = 0.

When T = 0 K, the Fermi surface EF with spin direction parallel to H is

&Ft = p+pgH (pg is the Bohr magneton) while the Fermi surface EF with

spin direction opposite to H is EFJ = p - pgH. Therefore, there exists a

net spin magnetic moment parallel to H. Hence n = 0.

2104

electrons in a "box" of volume V =

1 cm3. The walls of the box are infinitely high potential barriers. Calculate

the following within a factor of five and show the dependence on the relevant

physical parameters:

Take a system of N = 2 x

(a) the specific heat, C,

(b) the magnetic susceptibility, x,

(c) the pressure on the walls of the box, p,

(d) the average kinetic energy, (Ek).

( Cham g 0)

Solution:

The density of states in k space is given by

4rk2

86 dk'

D(k)dk = ZV. -