Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S16 / 1 / 2 / Solution

For total nuclear density $n$, $n_e=n\zeta$ and $n_H=n(1-\zeta )$. Saha's equation becomes $n{\zeta }^2/(1-\zeta )=C{T}^{3/2}{e}^{-R/(k_{B}T)}$. At $\zeta \ll 1$,The factor two in the exponent follows from taking a square root; using the full ionization temperature would double the predicted logarithmic slope.