 Solution

In a large volume, one one-particle state occupies momentum-space volume $h^3/V$. A spherical shell of radius $p$ has volume $4\pi p^{2}dp$, so for spin degeneracy $g$ the density of states is $4\pi gV{p}^{2}dp/h^3$. With ultra-relativistic dispersion $E=cp$ and zero chemical potential, the Bose-Einstein distribution givesThe convergent dimensionless integral is ${\pi }^4/15$, obtained by expansion into exponentials and summing $\sum n^{-4}$. Therefore $ϵ\propto T^4$ and $\alpha =4$. Photons are massless bosons with two polarizations; equilibrium emission and absorption do not conserve photon number and force ${\mu }_{\gamma }=0$.

For recombination, chemical equilibrium imposes ${\mu }_p+{\mu }_e={\mu }_H+{\mu }_{\gamma }={\mu }_H$, including rest energies in the chemical potentials. Divide the three nonrelativistic number-density expressions to obtainHere $m_p{c}^2+m_e{c}^2-m_H{c}^2=I$. Neglect the relative proton/hydrogen mass difference and take the ground-state degeneracy ratio $g_p{g}_e/g_H=1$ (electron and proton spins give $2\cdot 2$ and hydrogen gives $4$). Charge neutrality gives $n_p=n_e$. This yields the stated Saha equation

With $n_e=X_e{n}_B$ and $n_H=(1-X_e)n_B$, the ratio asked for isEquivalently the prefactor is $16\pi \zeta (3)\eta (2\pi {)}^{-3/2}[kT/(m_e{c}^2){]}^{3/2}$. Because photons vastly outnumber baryons, even a small high-energy tail can keep hydrogen ionized. Neutralization therefore requires exponential suppression of that tail far below $kT=I$, explaining the much lower recombination temperature. The equilibrium formula is an estimate; the actual expanding-universe recombination kinetics eventually depart from Saha equilibrium.