 Solution

Write $n_b=n_p+n_H$ for the cosmological baryon density in this hydrogen-only model, and $x_e=n_p/n_b$ for the hydrogen ionization fraction. By charge neutrality, $n_e=n_p$, so the Saha ionization equation becomesThe baryon-to-photon ratio is $\eta =n_b/n_{\gamma }$, and the equilibrium photon number density is $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$. At an order-one hydrogen ionization fraction, neglecting factors of order one therefore givesReplacing $T$ inside the slowly varying logarithm by $B$ givesWith $B=13.6\mathrm{eV}$, $M_e=5\times 10^5\mathrm{eV}$ and $\eta =10^{-10}$, this is $T\simeq 0.351\mathrm{eV}\simeq 4.1\times 10^3\mathrm{K}$. Solving the preceding implicit approximation instead gives $T\simeq 0.306\mathrm{eV}\simeq 3.5\times 10^3\mathrm{K}$. Both are reasonable estimates of the recombination temperature at the accuracy requested. The small baryon abundance delays hydrogen recombination: even when $T\ll B$, the high-energy tail of the abundant photons can still ionize the much rarer hydrogen atoms.