91Ó°ÊÓ

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 $\mathrm{Î\pm }$) equals the number of ionised hydrogen atoms plus the number of un-ionised hydrogen atoms:

$N_p+N_{\mathrm{H}}=\mathrm{Î\pm }{N}_{\mathrm{γ}}$

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

$N_{\mathrm{γ}}=8\mathrm{π}(2.404)V{\left(\frac{kT}{hc}\right)}^3$

Thus,

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

Substitute into (1)

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

Substitute int (3):

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

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

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

Let

$$\begin{gathered} \mathrm{β}=\mathrm{Î\pm }b{\left(\frac{kT}{hc}\right)}^3{v}_Q{e}^{I/kT}\text{where}{v}_Q={\left(\frac{h^2}{2\mathrm{Ï€}mkT}\right)}^{3/2} \\ \mathrm{β}=\mathrm{Î\pm }b{\left(\frac{kT}{2\mathrm{Ï€}m{c}^2}\right)}^{3/2}{e}^{I/kT} \end{gathered}$$

Therefore,

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

This is a quadratic equation.

The solution for quadratic equation is:

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

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

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

Then

$$\begin{gathered} \mathrm{β}=\mathrm{Î\pm }b{\left(\frac{It}{2\mathrm{Ï€}m{c}^2}\right)}^{3/2}{e}^{1/t} \\ \mathrm{β}=\mathrm{Î\pm }b{\left(\frac{I}{2\mathrm{Ï€}m{c}^2}\right)}^{3/2}{t}^{3/2}{e}^{1/t} \end{gathered}$$

Substitute the given values to find the constant:

$$\begin{gathered} \mathrm{β}=8\mathrm{Ï€}(2.404)\left(\mathrm{Î\pm }\right){\left(\frac{13.6×1.6×{10}^{-19}\mathrm{J}}{2\mathrm{Ï€}\left(9.11×{10}^{-31}\mathrm{kg}\right){\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)}^2}\right)}^{3/2}{t}^{3/2}{e}^{1/t} \\ \mathrm{β}=\left(5.245×{10}^{-7}\right)\left(\mathrm{Î\pm }\right)t^{3/2}{e}^{1/t} \end{gathered}$$

Substitute into (7)

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

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

The graph is: