= Solution
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 $m_H=m_p+m_e-\mathcal B$:
$$
\frac{n_en_p}{n_H}=\frac{g_eg_p}{g_H}\left(\frac{m_em_p}{m_H}\frac{T}{2\pi}\right)^{3/2}e^{-\mathcal B/T}.
$$
For ground-state <hydrogen>, including its spin states, $g_e=g_p=2$, $g_H=4$; hence the degeneracy factor is one. Since $m_p/m_H\simeq1$, the right side is $(m_eT/2\pi)^{3/2}e^{-\mathcal B/T}$.
Charge neutrality gives $n_p=n_e$, while baryon conservation gives $n_b=n_p+n_H$. Consequently $n_e=n_p=X_en_b$ 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 equation>
$$
\boxed{\frac{1-X_e}{X_e^2}=\frac{2\zeta(3)}{\pi^2}\eta\left(\frac{2\pi T}{m_e}\right)^{3/2}e^{\mathcal B/T}}.
$$
Writing $y=\mathcal B/T$, its coefficient is $(2\zeta(3)/\pi^2)\eta(2\pi\mathcal B/m_e)^{3/2}\simeq3.2\times10^{-16}$, consistent with the rounded $3\times10^{-16}$. The negligible mass-ratio correction and ground-state approximation are the assumptions behind this form.
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