Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S16 / 1

Assume dilute, nondegenerate ideal electrons, protons and neutral atoms in thermal and chemical equilibrium, with charge neutrality $n_p=n_e$. Set ${\mu }_H={\mu }_p+{\mu }_e$, using each ideal species' translational factor $n{\lambda }_{\mathrm{th}}^3/g$ and hydrogen ground-state binding energy $-R$. Neglect excited states and take the heavy proton/atom thermal-wavelength ratio as one. With electron spin and neutral-atom degeneracies treated in the convention of the source, the remaining factor isDifferent partition-degeneracy conventions must be propagated consistently. Screening, dense-plasma level dissolution and departures from local thermodynamic equilibrium lie outside this derivation.

For total nuclear density $n$, $n_e=n\zeta$ and $n_H=n(1-\zeta )$. Saha's equation becomes $n{\zeta }^2/(1-\zeta )=C{T}^{3/2}{e}^{-R/(k_{B}T)}$. At $\zeta \ll 1$,The factor two in the exponent follows from taking a square root; using the full ionization temperature would double the predicted logarithmic slope.

Write $u=1/T$. Then $\log \zeta =\mathrm{constant}-(3/4)\log u-T_{H}u$, giving local slope $d{\log }_{10}\zeta /du=-(T_H+3T/4)/\log 10$. At 5000–6000 K its magnitude is about $3.6\times 10^4$ K. The low-ionization tail in the original graph drops roughly one decade per $2.8\times 10^{-5}$ K${}^{-1}$, consistent with this value.

At $u=2.0\times 10^{-4}$ K${}^{-1}$, the graph gives $T\simeq 5000$ K and $\zeta \simeq 0.004$. Invert the full equation to get $n=C{T}^{3/2}{e}^{-R/(k_{B}T)}(1-\zeta )/{\zeta }^2$, yielding $\boxed{n\sim 10^{18}{\mathrm{m}}^{-3}}$. The neighboring tail markers give the same order; plotting precision and the squared dependence on the read ionization fraction preclude claiming many significant digits.