 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 givesThe 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. ThusThe 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.