 Solution

Assume dilute nondegenerate nonrelativistic particles, a common temperature, chemical equilibrium, charge neutrality $n_p=n_e$, and photons with zero chemical potential. The reaction balance gives ${\mu }_p+{\mu }_e={\mu }_H$. Dividing the two charged-species equilibrium densities by the neutral density cancels their chemical-potential factors and leavesFor ground-state hydrogen, $g_p=g_e=2$ and $g_H=4$, and $m_H\simeq m_p$ in the thermal prefactor. The rest-mass difference is $I/c^2$. These approximations give the Saha ionization equationExcited levels or different degeneracy conventions would change the prefactor; the stated assumptions give the printed normalization.

Write the right side as $S(T)$. Since $n_e=X_e{n}_B$ and $n_H=(1-X_e)n_B$, the hydrogen ionization fraction obeys $X_e^2/(1-X_e)=S(T)/n_B$. With $n_B=\eta n_{\gamma }$, the given blackbody photon density then yieldsThere are vastly more photons than baryons. Even when the typical photon energy is below the ionization energy, the high-energy tail can contain enough ionizing photons to keep nearly every baryon ionized. Neutral hydrogen forms only when that tail becomes sufficiently depleted relative to the very small baryon-to-photon ratio. This entropy/number-density effect explains why cosmological recombination occurs at a temperature far below $I$, rather than simply when $kT$ first falls below the binding energy. The Saha ionization equation describes the equilibrium estimate; late kinetic freeze-out is a separate limitation of that approximation.