Statistical Mechanics - Physics at Oregon State University

A.2. SOLUTIONS FOR CHAPTER 2. 239
we get
d e
dx
x − e−x ex + e−x = 1 − (ex − e−x ) 2
(ex + e−x )
CV = 4NB2
kT 2
1
2 =
(e B
B
kT + e− kT ) 2
4
(e x + e −x ) 2
For T to zero this approaches zero exponentially, for T to infinity this goes to
zero inverse quadratically, and the approaching zero confirms what we found in
(b).
We can write CV = Nkg( B
kT ) with
g(x) =
4x 2
(e x + e −x ) 2
g ′ 8x
(x) =
(ex + e−x ) 2 − 4x2 (ex − e−x )
(ex + e−x
2
= g(x)
) 3
which is zero if
0 = 2
x − ex − e−x ex + e−x x − ex − e −x
e x + e −x
which happens when x is somewhat larger than 2, or kT0 ≈ 1
4 (ɛ2 − ɛ1), more or
less as in (b).
Problem 6.
Each positive ion can be in one of four states. If the central atom is at the origin,
the x coordinates of these states are − 1
1
2a for i=1,2 and + 2a for i=3,4. The
component of the dipole along the x direction is − 1
1
2ea and + 2ea respectively,
and the energies are + 1
1
2eaE and − 2eaE. The state of atom j is therefore given
by a number sj which runs from 1 to 4, and the partition function is
Z(T ) =
s1,···,sN
1 −
e kT j ɛ(sj)
Again, the energies are independent and we get
Z(T ) = (Z1(T )) N
eaE
eaE
+
Z1(T ) = 2 e 2kT −
+ e 2kT
The probability of finding a state with quantum numbers s1, · · · , sN is therefore:
P rob(s1, · · · , sN ) = 1 1
e− kT j
Z(T ) ɛ(sj)