Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S16 / 3

Assume a dilute nondegenerate ideal mixture in thermal and chemical equilibrium, electrical neutrality $N_e=N_p$, and ground-state hydrogen only. Choose the zero of internal energy as a separated proton and electron, so the atom has energy $-R$. Its single-particle partition function contains $e^{R/(k_{\mathrm{B}}T)}$:Their ratio has one remaining power of $V$. Neglecting binding mass in $m_H\simeq m_p$ makes the translational mass factor $(m_e{m}_p/m_H{)}^{3/2}\simeq m_e^{3/2}$. For unresolved ground-state spin states, $g_e{g}_p/g_H=1$: including proton spin gives $2\times 2/4$, or omitting it consistently gives $2\times 1/2$. The given law of mass action therefore yieldsThis is the Saha equation with the degeneracy convention used by the source. Equivalently $n_e^2/n_H=(2\pi m_e{k}_{\mathrm{B}}T/h^2{)}^{3/2}{e}^{-R/(k_{\mathrm{B}}T)}$.

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.

Low excited states lie many thermal energies above the ground state at the inferred temperature, so their contributions to the atomic internal partition function are exponentially suppressed. For hydrogen, $E_2-E_1=10.2$ eV and the level degeneracy ratio is $g_2/g_1=4$. HenceThe ground-state approximation is therefore excellent for this level under the assumed thermal-equilibrium conditions. A formal sum over infinitely many isolated-atom Rydberg states needs a physical high-level cutoff in a plasma; it should not be used to overturn the finite-level Boltzmann estimate. Radiation-driven nonequilibrium populations would require a different model.