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Thermal photon number density is proportional to $T^3$. A sudden photon heating factor $(1+r{)}^{1/4}$ therefore raises photon number by $(1+r{)}^{3/4}$ and dilutes the baryon-to-photon ratio by its inverse when baryon number is conserved. Comparing Big Bang nucleosynthesis and Cosmic microwave background baryon-density determinations can test entropy injection between the two epochs, subject to changes in the expansion rate and nuclear abundances.

For chemical equilibrium $D\leftrightarrow n+p$, Maxwell--Boltzmann densities and ${\mu }_D={\mu }_n+{\mu }_p$ give, after taking $m_n\simeq m_p$ and $m_D\simeq 2m_p$,Writing $X_i=n_i/n_B$, $n_B=\eta n_{\gamma }$, and $n_{\gamma }=16\pi \zeta (3)(k_{B}T{)}^3/(hc{)}^3$ givesThe very small baryon-to-photon ratio suppresses deuterium until $T$ is far below the binding-energy scale.

At $X_e=1/2$, the left side of the Saha equation is two. With its rounded coefficient, the recombination temperature therefore satisfiesNeglecting the logarithmic prefactor gives $y\sim 36$ andThis is the requested rough scale. Retaining the prefactor and the unrounded logarithm gives $y\simeq 42.0$ and $T_{\mathrm{rec}}\simeq 0.32$ eV; the displayed $0.4$ eV is not a high-accuracy numerical root of the Saha equation.

The small baryon-to-photon ratio means that radiation and the entropy of the free charged particles strongly favor ionization. A low mean photon energy does not eliminate the energetic tail, and ionized particles have a large phase-space advantage. The exponential binding factor must become very large before neutral atoms dominate this dilute gas. Accordingly $\mathcal{B}/T$ is of order forty rather than order one, so recombination occurs well below the hydrogen binding energy.

The baryon-to-photon ratio is extremely small, of order $10^{-9}$. Thus even when the typical photon energy is below $I$, the high-energy tail of the blackbody radiation contains enough photons above the ionization energy to reionize almost every newly formed hydrogen atom. Cosmological recombination becomes effective only when this exponentially small tail is also small compared with the baryon-to-photon ratio. Consequently recombination occurs at a temperature much lower than $I/k$, consistent with $kT\simeq 0.03I$. The exact freeze-out history also depends on reaction rates and expansion; the equilibrium estimate explains the large suppression of the temperature.

Let $Y_{\mathrm{He}}\simeq 0.25$ be the present primordial helium mass fraction. If $n_{\mathrm{He}}$ is the helium-nucleus density and $n_H$ the hydrogen-nucleus density, thenSince $Y_{\mathrm{He}}=4n_{\mathrm{He}}/n_b$,This retains the correction from neutrons bound in helium. Charge neutrality requires one Electron per proton, including bound electrons, so with the baryon-to-photon ratio $\eta =n_b/n_{\gamma }\simeq 6\times 10^{-10}$,

In chemical equilibrium, the reaction $p+e\leftrightarrow H+\gamma$ obeys chemical-potential balance for ionizationbecause a photon gas in equilibrium has photon chemical potential ${\mu }_{\gamma }=0$. Applying the given nonrelativistic Maxwell-Boltzmann distribution to each massive species givesThe Hydrogen binding energy is $I=m_e+m_p-m_H$ in units with $c=1$. Thus Saha's equation isFor ground-state hydrogen, $g_e=g_p=2$ and $g_H=4$, so the degeneracy factor is one. Since $m_p\simeq m_H$, the translational factor is commonly written $(2\pi m_e{k}_{B}T/h^2{)}^{3/2}$.

Charge neutrality gives $n_e=n_p$. Write $n_B=n_e+n_H\simeq \eta n_{\gamma }$. Thenand henceCombining this with the stated photon density produces

In our universe the baryon-to-photon ratio is tiny, so there are roughly $10^9$ photons per baryon. Even when the mean photon energy is far below $I$, the high-energy blackbody radiation tail contains enough ionizing photons to suppress neutral hydrogen. The recombination temperature is reached only after the Boltzmann factor overwhelms this large photon-to-baryon ratio, at $k_{B}T\simeq 0.3\mathrm{eV}$.

For a universe with $\eta =1$, take the degeneracy factor as one, $m_H\simeq m_p$, and define $x=I/(k_{B}T)$. Neutral formation occurs when the boxed ratio is of order one, giving approximatelyWith $I=13.6\mathrm{eV}$ and $m_e=5.11\times 10^5\mathrm{eV}$, this gives $x\simeq 19$ and thereforeto order unity. This is much hotter than recombination in our universe because there is no enormous excess of photons, but it remains below $I$ because the electron's large translational phase space still favours the ionized state.