 Solution

Define fractional ionization as $x=n_p/(n_H+n_p)$ and use electrical neutrality. For $x=0.01$ and $n_e=10^9{\mathrm{m}}^{-3}$, $n_H=(1-x)n_e/x=99n_e$. Thus the density form of the Saha equation requiresSolving this monotone equation, conveniently after taking its logarithm, givesThe thermal energy is only about $2.2\%$ of the $13.6$ eV ionization energy. Ionization is not decided by setting $k_{\mathrm{B}}T=R$: the high-energy tail and the large translational state-count factor in chemical equilibrium allow a residual ionized fraction at a much lower temperature. This is the equilibrium estimate for the stipulated density and fraction, without inferring an expansion age from it.