Solution
= Solution
For total nuclear <density> $n$, $n_e=n\zeta$ and $n_H=n(1-\zeta)$. Saha's equation becomes $n\zeta^2/(1-\zeta)=C T^{3/2}e^{-R/(k_BT)}$. At $\zeta\ll1$,
$$
\boxed{\zeta\simeq\sqrt{C/n}\,T^{3/4}e^{-T_H/T},\quad m=\tfrac34,\quad T_H=\frac R{2k_B}\simeq7.89\times10^4\ \mathrm K.}
$$
The factor two in the exponent follows from taking a <square root>; using the full ionization <temperature> would double the predicted logarithmic slope.