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Chemical equilibrium imposes ${\mu }_e+{\mu }_p={\mu }_H+{\mu }_{\gamma }$. For blackbody photons, ${\mu }_{\gamma }=0$. Define the positive Hydrogen binding energy by $I=(m_e+m_p-m_H)c^2$, approximately $13.6\mathrm{eV}$. Multiplying the two Maxwell-Boltzmann distribution number densities and dividing by the neutral-hydrogen density givesUse $g_e{g}_p/g_H=1$ and $m_p/m_H\simeq 1$ to obtain the Saha ionization equationThe assumptions are nonrelativistic, dilute, nondegenerate species at a common temperature, ground-state hydrogen with the stated degeneracy factors, negligible mass correction in the prefactor, and sufficiently rapid reactions to maintain chemical equilibrium.

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.

Chemical equilibrium for $p+e^{-}\rightleftharpoons H+\gamma$ gives ${\mu }_p+{\mu }_e={\mu }_H$, because the photon chemical potential vanishes. Applying the nonrelativistic number-density formula to all three massive species and taking the ratio givesUse charge neutrality $n_p=n_e$, the Hydrogen binding energy $E_{\mathrm{bind}}=m_p{c}^2+m_e{c}^2-m_H{c}^2$, $m_H\simeq m_p$, and the ground-state statistical weights $g_H=g_p{g}_e$. ThenThe approximations are a dilute, nondegenerate ideal gas in thermal and chemical equilibrium, nonrelativistic electrons, protons and hydrogen, negligible excited hydrogen states, and negligible corrections from the proton-to-hydrogen mass ratio. This relation is the Saha ionization equation.

Write $n_b=n_e+n_H\simeq \eta n_{\gamma }$. Charge neutrality gives $n_p=n_e$, and thereforeSubstitution into the Saha ionization equation yieldsUsing the given photon density,

The baryon-to-photon ratio is extremely small. Even when $k_{B}T\ll I$, the high-energy tail of the enormous photon population contains enough ionizing photons per baryon to delay cosmological recombination. The exponential factor must become large enough to overcome both $\eta \sim 10^{-9}$ and the phase-space factor, which occurs near $k_{B}T\simeq 0.3\mathrm{eV}$ rather than near $I$.

With $X_e=n_e/n_b$, charge neutrality and $n_b=n_p+n_H$ giveMultiplying part a by $n_b=\eta n_{\gamma }$ and using $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$ yields the Saha ionization equationAlthough $T\ll B_H$ suppresses ionization of an individual atom, the baryon-to-photon ratio is only about $10^{-9}$. The enormous number of photons per baryon leaves enough photons in the high-energy thermal tail to ionize hydrogen until $T\simeq 0.3\mathrm{eV}$, far below $B_H$.

Chemical equilibrium for $p+e\leftrightarrow H$ givesand charge neutrality gives $n_p=n_e$. Insert the nonrelativistic Maxwell-Boltzmann number densities, use $m_H{c}^2=m_p{c}^2+m_e{c}^2-E_{\mathrm{bind}}$, and neglect the proton correction in the reduced mass. The spin degeneracies cancel in the ground-state approximation. One obtains Saha equationEquivalently, its reciprocal is the ionization ratio. The exponential favors neutral hydrogen as the temperature falls, while the translational phase-space factor favors ionization at high temperature.

In the nonrelativistic regime,The assumption $E-\mu \gg k_{B}T$ puts the Fermi-Dirac distribution in its Maxwell-Boltzmann limit, soTherefore the nonrelativistic Maxwell--Boltzmann number density is

For $p+e^{-}\leftrightarrow H+\gamma$, chemical equilibrium and the vanishing photon chemical potential implyApplying the number-density formula to each massive species and eliminating the chemical potentials givesUse charge neutrality, $n_p\simeq n_e$, the mass approximation $m_H\simeq m_p$, and the convention in the question that suppresses the order-one internal-degeneracy ratio $g_e{g}_p/g_H$. With the Hydrogen binding energywe obtain the Saha ionization equationThe assumptions are thermal equilibrium and chemical equilibrium before cosmological recombination, a nonrelativistic nondegenerate gas, zero photon chemical potential, charge neutrality, negligible proton-electron mass correction in the translational prefactor, and the stated degeneracy-factor approximation.

Chemical equilibrium for $e^{-}+p\leftrightarrow H+\gamma$ gives ${\mu }_e+{\mu }_p={\mu }_H$, since ${\mu }_{\gamma }=0$. Insert the supplied nonrelativistic equilibrium densities. Neglecting $m_e/m_p$ in the translational reduced mass and using the stated degeneracy convention givesCharge neutrality gives $n_p=n_e$, soThis is the hydrogen Saha ionization equation.