[You may work in units of the speed of light, so that $c=1 .$ ]

Consider the process where protons and electrons combine to form neutral hydrogen atoms;

$$p^{+}+e^{-} \leftrightarrow H^{0}+\gamma$$

Let $n_{p}, n_{e}$ and $n_{H}$ denote the number densities for protons, electrons and hydrogen atoms respectively. The ionization energy of hydrogen is denoted $I$. State and derive $S a h a$ 's equation for the ratio $n_{e} n_{p} / n_{H}$, clearly describing the steps required.

[You may use without proof the following formula for the equilibrium number density of a non-relativistic species $a$ with $g_{a}$ degenerate states of mass $m$ at temperature $T$ such that $k_{B} T \ll m$,

$$n_{a}=g_{a}\left(\frac{2 \pi m k_{B} T}{h^{2}}\right)^{3 / 2} \exp \left([\mu-m] / k_{B} T\right)$$

where $\mu$ is the chemical potential and $k_{B}$ and $h$ are the Boltzmann and Planck constants respectively.]

The photon number density $n_{\gamma}$ is given as

$$n_{\gamma}=\frac{16 \pi}{h^{3}} \zeta(3)\left(k_{B} T\right)^{3}$$

where $\zeta(3) \simeq 1.20$. Consider now the fractional ionization $X_{e}=n_{e} /\left(n_{e}+n_{H}\right)$. In our universe $n_{e}+n_{H}=n_{p}+n_{H} \simeq \eta n_{\gamma}$ where $\eta$ is the baryon-to-photon number ratio. Find an expression for the ratio

$$\frac{\left(1-X_{e}\right)}{X_{e}^{2}}$$

in terms of $k_{B} T, \eta, I$ and the particle masses. One might expect neutral hydrogen to form at a temperature given by $k_{B} T \sim I \sim 13 \mathrm{eV}$, but instead in our universe it forms at the much lower temperature $k_{B} T \sim 0.3 \mathrm{eV}$. Briefly explain why this happens. Estimate the temperature at which neutral hydrogen would form in a hypothetical universe with $\eta=1$. Briefly explain your answer.