We start then with the Boltzmann equation, which we can express:

N_{2 }/ N_{1 } = [g_{2} / g_{1 }]** _{ }**exp (- E2 – E1) / kT

That is, for the atoms of a given element
in a specified state of ionization, the ratio of the number of atoms N_{2 }with energy
E2, to the number of atoms N_{1} with energy E1, in different states of ionization
is given by the above formula. The same form of the equation can also be used
to find the ratio of probabilities, i.e. that the system will be found in any
of the g_{2 }degenerate states with energy E2 to the probability
that the system is in any of the g** _{1}**
degenerate states E1, viz.

P(E2)** _{ }**/ P(E1)

**= [g**

_{ }_{2}/ g

_{1 }]

**exp (- E2 – E1) / kT**

_{ }Thus, the
Boltzmann equation can be posed in two forms.
In statistical mechanics we also have the partition function:

Z
= å _{j} exp ( - e_{ j} )/ t

Which is just
the summation over the Boltzmann factor (exp ( - e_{ j} )/ t ) for all states j for which the number of particles (N) is
constant. We will find it useful to rewrite it:

Z = g_{1} + å^{¥} _{j = 2} g _{j} exp (- E_{ j} – E1) / kT

Of interest now are the relative numbers
of atoms in ionization stage i, which is written:

N
_{e} N _{i + 1}
/ N _{i} =

2 Z_{ i + 1} / Z _{i} (2 p** **m _{e}** **kT/ h ^{2}) ^{1.5} e ^{- }^{c}^{ i}^{/
kT}^{}

This is
the *Saha* equation, named after the Indian astrophysicist who first derived
it. Here, N _{e} is the number of free electrons per unit
volume and c_{ i} is the *ionization
potential* of the ith ionization stage. Thus, the equation relates the
number of atoms in two successive
ionization stages to the quantities that are relevant. As per our
introduction to quantum mechanics, the factor ‘2’ in the equation refers to the
two possible spins of the free hydrogen election with spin quantum number: m _{s}
= __+__½.

Recall that for thermodynamic equilibrium, the
rate of ionization cannot exceed the rate of recombination[1]. In other words, the rate at which atoms in
the ith stage are ionized (i.e. to the i +1st stage) must equal the rate at
wich ions in that i +1st stage are recombining with free electrons to form ions
in the ith stage. The latter depends on N _{e} N _{i + 1} and the former on N _{i} . Hence, Saha’s
equation simply expresses the fact these two processes must occur at the same
rate.

log(N
_{e} N _{i + 1}
/N _{i}) =

15.38
+ log (2 Z_{ i + 1}
/ Z _{i }) +
1.5 log T – 5040 c_{ i}/
T

The units here are important to note and
are consistent with the ionization potential being measured in electron volts
(eV). Therefore N _{e} must be
in *particles per cubic centimeter*.

Yet another way to express the Saha equation is to introduce the
electron pressure, P _{e} . This acknowledges that each separate
species of particle makes its own contribution to the total gas pressure. The
free electrons in a gas therefore produce a pressure given by: P _{e} =
N _{e} kT.

Then we may write another log form of the
Saha equation:

log(P
_{e} N _{i + 1}
/N _{i}) =

-0.48 + log (2 Z_{ i + 1} / Z _{i }) + 2.5 log T – 5040 c_{ i}/
T

* Example Problem*:

For a hydrogen plasma find:

(P
_{e} N _{i + 1}
/N _{i}) = P _{e}
N _{HII} /N _{HI}

at a temperature of 5040 K, given the hydrogen partition functions are:

Z_{ i + 1} = Z _{2} =
1 and:

Z
_{i} = Z _{2} =
2 with c_{ i }= 13.6 eV

* Solution*:

By
the Saha equation: log(P _{e} N
_{i + 1} /N _{i}) =

-0.48 + log (2 Z_{ i + 1} / Z _{i }) + 2.5 log T – 5040 c_{ i}/
T

= -0.48 + log (2) + 2.5 log (5040) – 5040 (13.6)/ 5040

= -0.48 + 0.30 + 9.25 – 13.6 = -4.83

Antilog
(-4.53) = 2.95 x 10 ^{-5}

A useful table that will come in handy for spectral line and ionization energy computations is presented below.

Careful inspection of the
table shows two quantum states with the
energy (-13.6 eV) and eight with (-3.40 eV). Thus, two states are degenerate for the
n=1 level and eight states are degenerate for the n=2 level. Since g _{n }= 2n^{2}, then:

At the n=1 level the statistical weight
is: g _{1 }= 2(1)^{2} =
2

At the n=2 level the statistical weight
is: g _{2 }= 2(2)^{2} = 8

*Suggested Problem:*

Consider a gas of neutral hydrogen (H).
Using the Boltzmann equation and the information in the table above, compute
the temperature at which one will expect e*qual numbers of atoms *in the ground
state and the first excited state.

[1] This must hold if the excitation and
ionization equations are assumed valid, hence the numbers of atoms in a given
level must not change with time.