558
10 Quantum Statistics
We may employ the Stirling approximation for the factorials in our expression for
the Helmholtz energy in order to obtain a simpler expression, namely
A −N de E D − k B T ln
N
N D
D
N
N de
de (N D − N de ) (N D −N de )
2
N de
,
(10.5.21)
and we may utilize this expression to obtain the chemical potential (Fermi level) μ
as
μ =
∂A
∂N de
N ce ,V ,T
= −E D − k B T
∂
∂N de
{N de ln 2 + N D ln N D − N de ln N de
−(N D − N de ) ln(N D − N de )}
= −E D − k B T {ln 2 − ln N de − 1 + 1 + ln(N D − N de )} ,
or
μ = −E D − k B T
ln 2 − ln
N de
N D − N de
.
(10.5.22)
We may also rewrite this result in the form
μ + E D
k B T
+ ln 2 = ln
N de
N D − N de
,
(10.5.23)
or, upon exponentiating both sides of the equation, in the form
N de
N D − N de
= 2e
β(μ+E D ) .
(10.5.24)
Upon rearranging this result, we obtain an expression for N de , viz.,
N de =
2N D e β(μ+E D )
1 + 2e β(μ+E D ) =
N D
1 +
1
2 e −β(E D +μ)
,
(10.5.25)
while we obtain an expression for N ce via relation (10.5.20):
N ce =
2
h 3
dp
e β((−μ) + 1
,
, =
p 2
2m ∗
ce
= 2
2πm ∗
ce k B T
h 2
3
2
e
βμ .
(10.5.26)