An Example

Consider a pure hydrogen gas at 10080 K and a gas pressure of $10^{-5}$ atmospheres. Find the ratio of the density of H$^+$ to H$_0$.

The ionization potential of hydrogen is $\chi_0=13.6$ eV. The ratio we want is

$$\frac{N_+}{N_0} = A (kT)^{3/2} N_e^{-1} \exp(-\chi_0/kT)$$

It is helpful here to work with $\log_{10}$ and rewrite the equation as

$$\log_{10}\frac{N_+}{N_0} = \log_{10}A + \frac{3}{2}\log_{10} ... ...}{2}\log_{10} T - \log_{10} N_e - \frac{\chi_0}{kT} \log_{10} e$$

In this equation $\chi_0$ would have to be in Joules, but substituting for $k$ and making the change in units to eV for $\chi_0$ gives

$$\log_{10}\frac{N_+}{N_0} = \log_{10}A + \frac{3}{2}\log_{10} ... ...rac{3}{2}\log_{10} T - \log_{10} N_e - \frac{5040}{T}\chi_0(eV)$$

In this situation every ionized H atom makes both a proton (H$^+$) and an electron $e$. The densities of these two components are equal

$$N_e = N_+$$

and consequently the Saha equation becomes

$$\log_{10}\frac{N_+}{N_0} = \log_{10}A + \frac{3}{2}\log_{10} ... ...rac{3}{2}\log_{10} T - \log_{10} N_e - \frac{5040}{T}\chi_0(eV)$$

$$\log_{10}\frac{N_+^2}{N_0} = \log_{10}A + \frac{3}{2}\log_{10} (1.3806 \times 10^{-23}) \;\;+$$

$$\frac{3}{2}\log_{10} (10080) - \frac{5040 \times 13.6}{10080}$$

Thus we have an equation involving the ion density $N_+$ and neutral density $N_0$

$$\log_{10}\frac{N_+^2}{N_0} = 20.588$$

$$\frac{N_+^2}{N_0} = 3.874\times10^{20}$$

We also know that the total pressure is due to neutral atoms, ions, and electrons. If $N$ is the total particle density, the equation of state at this temperature and pressure will be the ideal gas law

$$P = N k T$$

In this example

$$P = 10^{-5} \times 101325 \;\;\mathrm{pascal}$$

and

$$N = N_+ + N_e + N_0 = 2\;N_+ + N_0$$

At 10080 K the gas law gives

$$N = P / kT$$

$$N = \frac{1.01325}{1.3806 \times 10^{-23} \times 10080}$$

$$N = 7.281 \times 10^{18}\;\;\mathrm{m}^{-3}$$

$$2\;N_+ + N_0 = 7.281 \times 10^{18}\;\;\mathrm{m}^{-3}$$

We now have two equations and two unknowns:

$$\frac{N_+^2}{N_0} = 3.874\times10^{20}$$

$$2\;N_+ + N_0 = 7.281 \times 10^{18}\;\;\mathrm{m}^{-3}$$

We solve for $N_0$ in the second equation

$$N_0 = 7.281 \times 10^{18} - 2\;N_+$$

and substitute in the first one

$$\frac{N_+^2}{7.281 \times 10^{18} - 2\;N_+} = 3.874\times10^{20}$$

With rearrangement into a standard quadratic form

$$a N_+^2 + b N_+ + c = 0$$

$$a=1$$

$$b =7.748\times10^{20}$$

$$c = -2.82066 \times 10^{39}$$

the solutions are

$$N_+ = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$

The solution that is not negative is

$$N_+ = 3.62355 \times10^{18} \;\; \mathrm{m}^{-3}$$

Return now to the other equation

$$2\;N_+ + N_0 = 7.281 \times 10^{18}\;\;\mathrm{m}^{-3}$$

and substitute for the value of $N_+$ we have just found to get

$$N_0 = 7.281 \times 10^{18} - 2 \times 3.62355 \times10^{18}$$

$$N_0 = 3.39 \times 10^{16} \;\; \mathrm{m}^3$$

The ratio we were seeking is

$$\frac{N_+}{N_0} = \frac{3.62355 \times10^{18}}{3.39 \times 10^{16}}$$

$$\frac{N_+}{N_0} = 106$$

The ratio is

$$\frac{N_+}{N_0} = 106$$

At this temperature most of the hydrogen is ionized.