Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S16 / 2

At half ionization, electron and neutral densities are each $n_0/2$, so the left side of Saha's equation is $n_0/2$. Its logarithm changes by $d\log n_0=(3/2+R/(k_B{T}^{\prime }))d\log T^{\prime }$. To first order,This holds for small fractional changes and the supplied ground-state partition approximation, not an exact finite-difference identity.

The measured ratio is ten, so $R/(k_B{T}^{\prime })=8.5$. Taking $R=13.6$ eV gives $k_B{T}^{\prime }=1.60$ eV or $T^{\prime }\simeq 1.86\times 10^4$ K. Substitute into Saha's equation:Higher density favors recombination and reduces the entropy advantage of free electrons at a fixed ionized fraction, so a higher temperature is required. The ratio 8.5 means half ionization does not require $k_{B}T$ to equal the ionization energy: translational phase space and the exponential tail are important.

Hydrogen's $n=2$ excitation energy is $3R/4=10.2$ eV. With electronic degeneracy ratio $g_2/g_1=4$, its relative population is $4e^{-10.2/1.60}\simeq 6.8\times 10^{-3}$. Thus the first excited manifold contributes below one percent of the ground population at the inferred temperature, explaining the leading ground-state approximation.

There is a qualification to extending this argument to infinitely many Rydberg states: the isolated-atom partition sum with degeneracy $2n^2$ formally diverges. Plasma screening and level dissolution impose a physical cutoff. The small low-lying populations justify the simple approximation in its intended regime, not an unrestricted infinite partition sum.