192
4 Mean Values and Thermodynamics
p u
p o
=
1
e β(I p +μ) =
e −βI p
e βμ .
Because we have chosen to set up our reservoir essentially as a canonical ensemble
representing a mixture of N A atoms A, N A + ions A + , and N e − − 1 electrons, the
total pressure P for the mixture will be given by Dalton’s law of partial pressures,
i.e., P = P A + P A + + P e − . Moreover, the ratio of the probabilities p u and p o may
equivalently be represented as the ratio of the reservoir partial pressures P A + and
P A , so that we obtain the result
P A +
P A
=
e −βI p
e βμ .
If we now treat the bath electrons as a canonical ensemble of N e − electrons
(considered as an ideal gas, with electron spin neglected), we may determine the
chemical potential appearing in this expression via Eqs. (4.1.25c) and (3.2.21, 27)
as
μ ≡ μ e − = −k B T ln
V
N e − 3
e −
= −k B T ln
k B T
P e − 3
e −
,
with N e − N A + N A , so that the electrons can be treated as an ideal gas, with
P e − V e − = N e − k B T . In this approximation, e βμ is given by
e
βμ
=
P e − 3
e −
k B T
.
With this result, the ratio of the partial pressures is given by
P A + P e −
P A
=
k B T
3
e −
e
−I p /(k B T ) ,
which is one form for the Saha equation employed in plasma physics to compute the
degree of ionization of atoms in a weak plasma (see also Example 9.3 of Chap. 9,
Sect. 9.3).
Example 4.5 Effect of an impurity atom in a semiconductor crystal.
For an impurity atom/ion in a semiconductor crystal, including electron spin
degeneracy, we have three possible states: one ionized state (with no electron),
plus two un-ionized states (with one electron, either spin-up or spin-down) present.
Similar to what we have seen in the previous example, the ionized state (here the
bottom of the conduction band) is assigned energy = 0 and number N = 0, and
the un-ionized doubly degenerate state is assigned energy = −I p and number
N = 1, giving Gibbs factors e 0 = 1 for the first state, and e β(I p +μ) for the second
state.
4 Mean Values and Thermodynamics
p u
p o
=
1
e β(I p +μ) =
e −βI p
e βμ .
Because we have chosen to set up our reservoir essentially as a canonical ensemble
representing a mixture of N A atoms A, N A + ions A + , and N e − − 1 electrons, the
total pressure P for the mixture will be given by Dalton’s law of partial pressures,
i.e., P = P A + P A + + P e − . Moreover, the ratio of the probabilities p u and p o may
equivalently be represented as the ratio of the reservoir partial pressures P A + and
P A , so that we obtain the result
P A +
P A
=
e −βI p
e βμ .
If we now treat the bath electrons as a canonical ensemble of N e − electrons
(considered as an ideal gas, with electron spin neglected), we may determine the
chemical potential appearing in this expression via Eqs. (4.1.25c) and (3.2.21, 27)
as
μ ≡ μ e − = −k B T ln
V
N e − 3
e −
= −k B T ln
k B T
P e − 3
e −
,
with N e − N A + N A , so that the electrons can be treated as an ideal gas, with
P e − V e − = N e − k B T . In this approximation, e βμ is given by
e
βμ
=
P e − 3
e −
k B T
.
With this result, the ratio of the partial pressures is given by
P A + P e −
P A
=
k B T
3
e −
e
−I p /(k B T ) ,
which is one form for the Saha equation employed in plasma physics to compute the
degree of ionization of atoms in a weak plasma (see also Example 9.3 of Chap. 9,
Sect. 9.3).
Example 4.5 Effect of an impurity atom in a semiconductor crystal.
For an impurity atom/ion in a semiconductor crystal, including electron spin
degeneracy, we have three possible states: one ionized state (with no electron),
plus two un-ionized states (with one electron, either spin-up or spin-down) present.
Similar to what we have seen in the previous example, the ionized state (here the
bottom of the conduction band) is assigned energy = 0 and number N = 0, and
the un-ionized doubly degenerate state is assigned energy = −I p and number
N = 1, giving Gibbs factors e 0 = 1 for the first state, and e β(I p +μ) for the second
state.