 Saha baryon-abundance normalization

For a hydrogen-only gas with $g_e=g_p=2$, $g_H=4$ and $m_H\simeq m_p$, the Saha equation is $(1-X_e)/X_e^2=n_B[2\pi /(m_{e}T){]}^{3/2}{e}^{I/T}$. Defining ${\eta }_{\gamma }=n_B/n_{\gamma }$ and using $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$ gives coefficient $2\zeta (3){\eta }_{\gamma }/{\pi }^2$ multiplying $(2\pi T/m_e{)}^{3/2}{e}^{I/T}$. If instead ${\eta }_s=n_B/s$, with entropy density $s=(2{\pi }^2/45)g_{\ast s}{T}^3$, that coefficient is $(2{\pi }^2/45)g_{\ast s}{\eta }_s$. The two expressions agree because $s/n_{\gamma }={\pi }^4{g}_{\ast s}/[45\zeta (3)]$. The reference entropy sector must be specified, especially after particles have decoupled.