= Solution
Since $n_e=n_p=X_en_b$ and $n_H=(1-X_e)n_b$,
$$
\frac{1-X_e}{X_e^2}=n_b\frac{n_H}{n_e^2}.
$$
Using $n_b=\eta n_\gamma$ and $n_\gamma=2\zeta(3)T^3/\pi^2$ yields
$$
\boxed{
\frac{1-X_e}{X_e^2}
=\frac{2\zeta(3)}{\pi^2}\eta
\left(\frac{2\pi T}{m_e}\right)^{3/2}e^{B_H/T}}.
$$
<Cosmological 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.
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