= Solution
Assume a dilute nondegenerate ideal mixture in thermal and chemical equilibrium, electrical neutrality $N_e=N_p$, and ground-state hydrogen only. Choose the zero of internal energy as a separated proton and electron, so the atom has energy $-R$. Its single-particle partition function contains $e^{R/(k_{\rm B}T)}$:
$$
Z_{{\rm sp},e}=g_eV\left(\frac{2\pi m_ek_{\rm B}T}{h^2}\right)^{3/2},\quad
Z_{{\rm sp},p}=g_pV\left(\frac{2\pi m_pk_{\rm B}T}{h^2}\right)^{3/2},
$$
$$
Z_{{\rm sp},H}=g_HV\left(\frac{2\pi m_Hk_{\rm B}T}{h^2}\right)^{3/2}e^{R/(k_{\rm B}T)}.
$$
Their ratio has one remaining power of $V$. Neglecting binding mass in $m_H\simeq m_p$ makes the translational mass factor $(m_em_p/m_H)^{3/2}\simeq m_e^{3/2}$. For unresolved ground-state spin states, $g_eg_p/g_H=1$: including proton spin gives $2\times2/4$, or omitting it consistently gives $2\times1/2$. The given law of mass action therefore yields
$$
\boxed{\frac{N_e^2}{N_H}=V\left(\frac{2\pi m_ek_{\rm B}T}{h^2}\right)^{3/2}e^{-R/(k_{\rm B}T)}.}
$$
This is the <Saha equation> with the degeneracy convention used by the source. Equivalently $n_e^2/n_H=(2\pi m_ek_{\rm B}T/h^2)^{3/2}e^{-R/(k_{\rm B}T)}$.
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