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In hydrogen-only cosmology, charge neutrality gives $n_e=n_p$ and baryon conservation gives $n_b=n_p+n_H$. The fraction $X_e$ lies between zero and one, with $n_H=(1-X_e)n_b$. With helium present, Electron abundance per hydrogen nucleus and per total baryon are different conventions and must be specified. The hydrogen-only Saha equation gives the equilibrium fraction.

For the reaction $p+e\leftrightarrow H+\gamma$, chemical equilibrium requires ${\mu }_p+{\mu }_e={\mu }_H$ because thermal photons have zero chemical potential. Insert the nonrelativistic Maxwell-Boltzmann distribution number densities and use $m_H=m_p+m_e-\mathcal{B}$:For ground-state hydrogen, including its spin states, $g_e=g_p=2$, $g_H=4$; hence the degeneracy factor is one. Since $m_p/m_H\simeq 1$, the right side is $(m_{e}T/2\pi {)}^{3/2}{e}^{-\mathcal{B}/T}$.

Charge neutrality gives $n_p=n_e$, while baryon conservation gives $n_b=n_p+n_H$. Consequently $n_e=n_p=X_e{n}_b$ and $n_H=(1-X_e)n_b$. Inverting the preceding ratio and inserting the photon number density with $n_b=\eta n_{\gamma }$ gives the hydrogen-only Saha equationWriting $y=\mathcal{B}/T$, its coefficient is $(2\zeta (3)/{\pi }^2)\eta (2\pi \mathcal{B}/m_e{)}^{3/2}\simeq 3.2\times 10^{-16}$, consistent with the rounded $3\times 10^{-16}$. The negligible mass-ratio correction and ground-state approximation are the assumptions behind this form.

DefineUsing $n_e=n_B\mathcal{F}$, $n_{j^{+}}=X_{j^{+}}{n}_j$, and $n_{j^0}=(1-X_{j^{+}})n_j$, the two Saha relations become the closed systemwhereThese are two equations for the two ionization fractions.

When $Y_p\to 0$, $\mathcal{F}=X_{H^{+}}$, and the hydrogen equation reduces tothe usual hydrogen-only Saha equation.