 Solution

For hydrogen, the electronic ground state has $g_{\mathrm{H}}=2$, while the proton has no electronic degeneracy, $g_{{\mathrm{H}}^{+}}=1$. For sodium, the neutral $3s{}^2{S}_{1/2}$ ground state has $g_{\mathrm{Na}}=2$ and the closed-shell ion has $g_{\mathrm{N}{\mathrm{a}}^{+}}=1$. Ignoring hyperfine structure and consistently cancelling nuclear-spin factors, $2g_{{\mathrm{M}}^{+}}/g_{\mathrm{M}}=1$ for both.

At $5000\mathrm{K}$,It is convenient to calculate the density $D=K_c{c}^{\circ }$, which avoids choosing an unnecessary numerical value of the standard density. The results areThus hydrogen is about $0.00409\%$ ionized, whereas sodium is about $53.6\%$ ionized in the stated ground-state ideal-gas model. The large difference is dominated by the smaller sodium ionization energy in the exponential. The weak-ionization approximation is excellent for hydrogen but would be inaccurate for sodium, so the quadratic solution is needed there.