9.3 Four Extended Examples
495
from which we obtain the value 0.0113 for K P (T = 1073 K). In this case, we
do have an experimental value, K P = 0.0114, with which we may compare our
computed result. If we recall that many of the approximations that we have utilized
in obtaining our calculated result for K P are worse the higher we go in temperature,
this improved level of agreement between theory and experiment indicates that it is
indeed likely that much of the difference between our calculated value at 1073 K and
the experimental value is due to the approximations that we have made in obtaining
our calculated result.
Example 9.3 Ionization of an alkali metal atom.
Consider the ionization of an alkali metal atom M to give a free electron plus the
M + ion,
M
2
S 1
2
I p
→ M
+
1 S 0
+ e
− ,
in which I p is the ionization potential for the reaction. The free electron is a spin1
2 fermion which has only translational energy; M and M + will have nuclear spin
degeneracies in common, and if we ignore the difference in mass between M and
M + , the translational partition functions for M and M + will cancel out in the
expression
K P =
z ◦
M + z ◦
e −
z ◦
M
1
N 0
e
−βI p
(9.3.14)
for K P . We then obtain
K P =
2(2πm e k B T )
3
2 V
◦
h 3
z el,M +
z el,M
1
N 0
e
−βI p .
(9.3.15)
Now, since ω el (M) = 2, and ω el (M + ) = 1, while V
◦ = N 0 k B T /P ◦ , we have
K P =
(2πm e k B T )
3
2
h 3
k B T
P ◦ e
−βI p
=
1
h 3 (2π)
3
2 k B
5
2 m e
3
2 T
5
2 e
−βI p
(15.75)(7.08 × 10 −58 )
2.909 × 10 −100 P ◦ m
3
2
e T
5
2 e
−βI p ,
(9.3.16)