Problems and Solutions on Thermodynamics and Statistical Mechanics

Statistical Phyaica 233

The result is shown in Fig. 2.15

(c) With E = - P2 we have N= -

2m

or

dx

ex-%, - 1 '

2T

h3

N/V= - (2mkT)3/2

where xcl. = p/kT 5 0. As N/V remains unchanged when T decreases,
p(T) increases and approaches zero.
(d) Let n be the number density and T, the critical temperature. Note
that at temperature T, the chemical potential p is near to zero and the
particle number of the ground state is still near to zero, so that we have

= -(2mkT,)3/2 2T

h3

where the integral

= 1.3066.

Hence

T--

(e) For bosons, p < 0. When T 5 T,, p w 0 and we have

n,>o =

exp(iEji;) - 1 '