Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 310 / 2 / b / Solution

With $X_e=n_e/n_b$, charge neutrality and $n_b=n_p+n_H$ giveMultiplying part a by $n_b=\eta n_{\gamma }$ and using $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$ yields the Saha ionization equationAlthough $T\ll B_H$ suppresses ionization of an individual atom, the baryon-to-photon ratio is only about $10^{-9}$. The enormous number of photons per baryon leaves enough photons in the high-energy thermal tail to ionize hydrogen until $T\simeq 0.3\mathrm{eV}$, far below $B_H$.