In the text I claimed that the universe was filled with ionised gas until its temperature cooled to about 3000 K. To see why, assume that the universe contains only photons and hydrogen atoms, with a constant ratio of 10⁹ photons per hydrogen atom. Calculate and plot the fraction of atoms that were ionised as a function of temperature, for temperatures between 0 and 6000 K. How does the result change if the ratio of photons to atoms is 10⁸ or 10¹⁰? (Hint: Write everything in terms of dimensionless variables such as t = kT/I, where I is the ionisation energy of hydrogen.)

The universe was filled with ionised gas until its temperature cooled to about 3000K.

Assume that the universe contains only photons and hydrogen atoms, with a constant ratio of 10⁹ photons per hydrogen atom.

If f is the fraction of atoms that are ionised, then the proportion of ionised to non-ionised atoms is:

$f=\frac{N_p}{N_p+N_{\mathrm{H}}}\left(1\right)$

Where N_p is number of protons and these are ionised hydrogen.

The equation is:

$\frac{N_p{N}_e}{N_{\mathrm{H}}}=\frac{V}{v_Q}{e}^{-I/kT}$

For each proton there is an electron so $N_e=N_p$, hence the equation becomes:

$\frac{N_p^2}{N_{\mathrm{H}}}=\frac{V}{v_Q}{e}^{-I/kT}\left(2\right)$

The number of photographs multiplied by a factor (say $\alpha$) equals the number of ionised hydrogen atoms plus the number of un-ionised hydrogen atoms:

$N_p+N_{\mathrm{H}}=\alpha N_{\gamma }$

Where $N_{\gamma }$is the number of photons and is given by:

$N_{\gamma }=8\pi (2.404)V{\left(\frac{kT}{hc}\right)}^3$

Thus,

$N_p+N_{\mathrm{H}}=\alpha bV{\left(\frac{kT}{hc}\right)}^3\left(3\right)$

Substitute into (1)

$$\begin{gathered} N_p=f\left(N_p+N_{\mathrm{H}}\right) \\ {N}_p=f\alpha bV{\left(\frac{kT}{hc}\right)}^3\left(4\right) \end{gathered}$$

Substitute int (3):

$$\begin{gathered} f\alpha bV{\left(\frac{kT}{hc}\right)}^3+N_{\mathrm{H}}=\alpha bV{\left(\frac{kT}{hc}\right)}^3 \\ {N}_{\mathrm{H}}=(1-f)\alpha bV{\left(\frac{kT}{hc}\right)}^3 \end{gathered}$$

Substitute from (5) and (4) into (2):

$$\begin{gathered} \frac{{\left(f\alpha bV{\left(\frac{kT}{hc}\right)}^3\right)}^2}{(1-f)\alpha bV{\left(\frac{kT}{hc}\right)}^3}=\frac{V}{v_Q}{e}^{-I/kT} \\ \frac{f\alpha b{\left(\frac{kT}{hc}\right)}^3}{(1-f)}=\frac{1}{v_Q}{e}^{-I/kT} \\ {f}^2\alpha b{\left(\frac{kT}{hc}\right)}^3{v}_Q{e}^{I/kT}=1-f\left(6\right) \end{gathered}$$

Let

$$\begin{gathered} \beta =\alpha b{\left(\frac{kT}{hc}\right)}^3{v}_Q{e}^{I/kT}\text{where}{v}_Q={\left(\frac{h^2}{2\pi mkT}\right)}^{3/2} \\ \beta =\alpha b{\left(\frac{kT}{2\pi m{c}^2}\right)}^{3/2}{e}^{I/kT} \end{gathered}$$

Therefore,

$$\begin{gathered} \beta f^2=1-f \\ \beta f^2+f-1=0 \end{gathered}$$

This is a quadratic equation.

The solution for quadratic equation is:

$f=\frac{-1+\sqrt{1+4\beta }}{2\beta }\left(7\right)$

Since we need to plot this equation, we need to change the variables into dimensionless variables, substituting the values

$t=\frac{kT}{I}$

Then

$$\begin{gathered} \beta =\alpha b{\left(\frac{It}{2\pi m{c}^2}\right)}^{3/2}{e}^{1/t} \\ \beta =\alpha b{\left(\frac{I}{2\pi m{c}^2}\right)}^{3/2}{t}^{3/2}{e}^{1/t} \end{gathered}$$

Substitute the given values to find the constant:

$$\begin{gathered} \beta =8\pi (2.404)\left(\alpha \right){\left(\frac{13.6\times 1.6\times {10}^{-19}\mathrm{J}}{2\pi \left(9.11\times {10}^{-31}\mathrm{kg}\right){\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2}\right)}^{3/2}{t}^{3/2}{e}^{1/t} \\ \beta =\left(5.245\times {10}^{-7}\right)\left(\alpha \right)t^{3/2}{e}^{1/t} \end{gathered}$$

Substitute into (7)

$f=\frac{-1+\sqrt{1+4\left(5.245\times {10}^{-7}\right)\left(\alpha \right)t^{3/2}{e}^{1/t}}}{2\left(5.245\times {10}^{-7}\right)\left(\alpha \right)t^{3/2}{e}^{1/t}}$

This formula can be plot $\text{first for}\alpha ={10}^{-9}\text{then for}{10}^{-8}\text{and}{10}^{-10}$on same graph using python. The code is:

The graph is: