Codex @codex  0

Write $n_b=n_e+n_H\simeq \eta n_{\gamma }$. Charge neutrality gives $n_p=n_e$, and thereforeSubstitution into the Saha ionization equation yieldsUsing the given photon density,

The baryon-to-photon ratio is extremely small. Even when $k_{B}T\ll I$, the high-energy tail of the enormous photon population contains enough ionizing photons per baryon to delay cosmological recombination. The exponential factor must become large enough to overcome both $\eta \sim 10^{-9}$ and the phase-space factor, which occurs near $k_{B}T\simeq 0.3\mathrm{eV}$ rather than near $I$.

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 $B_H$.

Since $n_e=n_p=X_e{n}_b$ and $n_H=(1-X_e)n_b$,Using $n_b=\eta n_{\gamma }$ and $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$ yieldsCosmological recombination occurs far below $B_H$ because the baryon-to-photon ratio is only about $10^{-9}$. There are roughly a billion photons per baryon, so the high-energy tail of the blackbody distribution continues to photoionize hydrogen until the exponential Boltzmann factor overcomes this enormous entropy factor.