The Saha-Boltzmann equation
Saha's ionization equation
Boltzman's excitation equation
The combined Saha-Boltzmann equation
The strength of spectral lines
The spectral sequence
One of the goals of spectroscopy is to obtain the abundance of the elements in a plasma, the chemical composition. Since the spectra of elements ions and molecules are quite unique, a qualitative analysis of a plasma is easy. But a quantitative analysis, the actual number density of a particular species of atom in a plasma, is a very difficult procedure. The qualitative analysis is simply the presence or absence of characteristic spectra, the quantitative analysis requires an interpretation of the intensity of characteristic lines in the spectra, and this may involve a second goal of spectroscopy, the determination of the actual state (temperature T, density r) of the plasma.

In the following discussion the temperature and density are assumed known and the number density of element Fe is to be found:

The strength of an absorption line depends on its intrinsic strength (Einstein A or Landenberg f-value) and the number of atoms in the (lower) level associated with the line. If the intrinsic strength of a line of Fe is known, the problem reduces to the number of atoms in the lower level which in turn is related to the number density of Fe, the thing we are after. The trick is to find a relation between the number density of Fe and the number in the lower level.

If the plasma is in thermodynamic equilibrium the problem is quite straightforward. For a given condition (T,density) a fraction of the atoms will be neutral, a fraction ionized, a fraction doubly ionized:

N(Fe)=N(Fe I)+N(Fe II)+N(Fe III)+...

The ratios N(Fe II)/N(Fe I), N(Fe III)/N(Fe II), and so on are given by Saha's ionization equation. For the rth ionization stage,

where U is the partition function, I the ionization potential (eV), Q=5040/T and Pe the electron pressure (dyne/cm2). (U and I are found in tables.)

Now at any given (T,density) most of the Fe will be in one of the stages so we have

Nr/N=Nr/(N1+N2+N3+...)=1/(N1/Nr+N2/Nr+N3/Nr+...)
which gives the fraction of atoms in the rTH stage of ionization. In the Sun most metals are in the first stage of ionization and the remainder neutral so we get N(Fe I)/N(Fe)=1/(1+N(Fe II)/N(Fe I)). The third term N(Fe III)/N(Fe I) and higher terms can be neglected.

Now we have to work out what fraction of the ion fraction is in the lower level involved in the spectral line. In thermodynamic equilibrium this is given by Boltzmann's excitation equation,

Where Nr,s is what we are after, the number of atoms in the r stage of ionization and s state of excitation, (Nr is the sum of all of the Nr,s), gr,s is the statistical weight of the level (2J+1) and cr,s is the excitation potential in eV (=8066/level energy in cm-1).

So

Nr,s=(Nr,s/Nr) (Nr/N) N,
the number of atoms capable of absorbing the line is the number of atoms (the answer) times the Boltzmann factor times the Saha factor. Not really all that bad!
But What Does It All Mean?
The Saha equation tells us which ionic species dominate. The higher the temperature, the higher the ionic species. In the Sun we find metals are mostly singly ionized, a small fraction is neutral and there are essentially no doubly or higher ion species present.

The Boltzmann equation tells us that, for a particular ionic species, the ground state (no excitation, C=0eV) is always the most populated and as the temperature rises the excited states become more and more likely to be populated. Lines arising from the ground level are generally the strongest lines observed and are called resonance lines.

Here is the result of applying the Saha equation to the relatively hard to ionize nitrogen atom with unit electron pressure, not a bad value for the main sequence:

Hey! There was a II after the N on the RED graph...oh well, the red squares give the N II/N population. Note how one or at most two ionization stages are all you need to consider in the calculations. Wanna see the actual data used to make the graph? Here is all 7k of it!
The Spectral Sequence
Consider the H atom for which I=13.595eV, the first excited level C1,2=10.2eV, gr,s=2, U1 is about 2, U2 is 1. In the Solar photosphere the temperature is roughly 5040K or Q=1 and Pe is between 10 and 100 so from Saha's equation we see H is completely neutral and we should observe lines of neutral H. Which lines? The resonance lines are Lyman lines and the longest wavelength La lies in the far UV (121.5nm). The only visible lines, the Balmer lines, arise from the first excited level for which the Boltzmann factor is 10-10.2 so fewer than one in ten billion H atoms participate. The Balmer spectrum is pretty weak in the Sun.

In Sirius, a typical A0V star, the temperature is roughly 10080 or Q=1/2 so from Saha's equation we see H is about equal parts H I and H II. There is less H I now but the Boltzmann factor is now 10-5.1 or about five orders of magnitude more than in the sun. The Balmer lines are quite strong in Sirius and relatively weak in the Sun, a typical G2V star.

As T rises above 10,000K the Boltzmann factor favours the Balmer lines more and more but the Saha ionization equation reveals a larger and larger H II/H I ionization fraction. By 16000K the amount of H I is quite small so for stars much hotter than B0 the Balmer spectrum is weak if at all visible. Even the Lyman spectrum vanishes.

By the way, not only do the Balmer lines peak at A0, so does the "Balmer Jump", the region around 350nm where H atoms in the n=2 (first excited) state can be ionized. The wavelength region between about 100nm and 350nm is often called the Balmer Continuum.