= Solution
<Chemical equilibrium> for $p+e^-\leftrightarrow H+\gamma$ implies $\mu_H=\mu_p+\mu_e$, because equilibrium <photons> have zero <chemical potential>. Dividing the three nonrelativistic Maxwell-Boltzmann <number densities> therefore gives
$$
\frac{n_H}{n_pn_e}=\frac{g_H}{g_pg_e}\left(\frac{m_H}{m_pm_e}\right)^{3/2}\left(\frac{2\pi}T\right)^{3/2}\exp\!\left(\frac{m_p+m_e-m_H}T\right).
$$
The <Hydrogen binding energy> is $I=m_p+m_e-m_H=13.6\,\mathrm{eV}$ in <natural units>. For ground-state <hydrogen>, with hyperfine splittings unresolved, $g_H=4$ and $g_p=g_e=2$, so the degeneracy ratio is one. Equivalently one can omit nuclear-spin degeneracy consistently from both the <proton> and atom. Excited <hydrogen> levels are exponentially suppressed at <recombination temperatures>. Also $m_H/m_p=1+O(m_e/m_p)$, so its effect in the prefactor is negligible. Thus
$$
\boxed{\frac{n_H}{n_pn_e}\simeq\left(\frac{2\pi}{m_eT}\right)^{3/2}e^{I/T}.}
$$
The sign of the exponential follows from the smaller <rest mass> of the bound atom: cooling favours neutral <hydrogen> once translational <entropy> no longer compensates the binding <energy>.
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