Quantum Statistics
247
the Pauli principle forbids more then one particle in each state). Thus, the two
distributions are equally probable.
The number of distributions can be verified by Eqs. (7.4), (7.7) and (7.10).
Example 2
The extension of the classical distributions for the case of bound and ionized
atoms in equilibrium is of special interest in astrophysics and plasma physics.
The equilibrium distribution is obtained by using arguments similar to those
used in Sec. 6.4 for obtaining the Einstein coefficients A and B.
The transitions in this case are
M 0
+
+
M
e
(7.129)
where M 0 is the neutral atom and M
+
is its ion. In contrast to the discussion in
Sec. 6.4, here the final states form a continuum. Let N 0 be the number of M 0
atoms and N + be the number of M
+
ions. The number of ionization transitions to
a set of states f i , is obtained from Eq. (6.60) as
N mn = B mn u(ω) N 0 f i
f i =
3/2
1/2
2 3
(2 )
4
m V
d
ε
ε
π
(7.130)
where f i is given in Eq. (7.29) and ω = ε + E I , E I being the ionization energy
and ε is the energy of the electron. The number of reverse reactions, i.e.
recombinations, is given by the first equation in Eq. (6.60) except that now the
expression is also proportional to the number dN e of electrons in f i states
N mn = ( B nm u (ω) + A nm ) N + dN e |
(7.131)
Equating N mn and N nm, gives for u (ω)
u (ω) =
0
nm
mn i
nm
e
A
N
B f
B
N dN
+
−
(7.132)
In analogy with Eq. (6.66) B nm is taken to be equal to B nm (this can be justified
by more rigorous arguments). Comparing u (ω) with the expression in Eq. (66.5)
gives
0
e
N dN
N
+
= exp [– (ε + E I )/kT] f i
(7.133)
Substituting for f i and integrating over dN e and ε, finally gives
0
e
n n
n
+
=
3/2
2
2
exp (– / )
I
mkT
E kT
h
π
(7.134)
where n 0 = N 0 /V, etc. This equation is known as the Saha equation (1920).
247
the Pauli principle forbids more then one particle in each state). Thus, the two
distributions are equally probable.
The number of distributions can be verified by Eqs. (7.4), (7.7) and (7.10).
Example 2
The extension of the classical distributions for the case of bound and ionized
atoms in equilibrium is of special interest in astrophysics and plasma physics.
The equilibrium distribution is obtained by using arguments similar to those
used in Sec. 6.4 for obtaining the Einstein coefficients A and B.
The transitions in this case are
M 0
+
+
M
e
(7.129)
where M 0 is the neutral atom and M
+
is its ion. In contrast to the discussion in
Sec. 6.4, here the final states form a continuum. Let N 0 be the number of M 0
atoms and N + be the number of M
+
ions. The number of ionization transitions to
a set of states f i , is obtained from Eq. (6.60) as
N mn = B mn u(ω) N 0 f i
f i =
3/2
1/2
2 3
(2 )
4
m V
d
ε
ε
π
(7.130)
where f i is given in Eq. (7.29) and ω = ε + E I , E I being the ionization energy
and ε is the energy of the electron. The number of reverse reactions, i.e.
recombinations, is given by the first equation in Eq. (6.60) except that now the
expression is also proportional to the number dN e of electrons in f i states
N mn = ( B nm u (ω) + A nm ) N + dN e |
(7.131)
Equating N mn and N nm, gives for u (ω)
u (ω) =
0
nm
mn i
nm
e
A
N
B f
B
N dN
+
−
(7.132)
In analogy with Eq. (6.66) B nm is taken to be equal to B nm (this can be justified
by more rigorous arguments). Comparing u (ω) with the expression in Eq. (66.5)
gives
0
e
N dN
N
+
= exp [– (ε + E I )/kT] f i
(7.133)
Substituting for f i and integrating over dN e and ε, finally gives
0
e
n n
n
+
=
3/2
2
2
exp (– / )
I
mkT
E kT
h
π
(7.134)
where n 0 = N 0 /V, etc. This equation is known as the Saha equation (1920).