Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S16 / 1

Assume ideal, nondegenerate particles in local thermodynamic equilibrium, ground-state atomic hydrogen, no molecules, and a common temperature. Let ${\lambda }_j=h/\sqrt{2\pi m_j{k}_{B}T}$ be the thermal de Broglie wavelength. Choose zero energy for a separated stationary proton and electron, so a neutral atom has ground-state energy $-R$. The classical chemical potentials areChemical equilibrium for $H\rightleftharpoons p+e$ requires ${\mu }_H={\mu }_p+{\mu }_e$. Rearranging givesIncluding electron and proton spin degeneracy, $g_e=2$, $g_p=2$ and $g_H=4$ when hyperfine splittings are neglected. Hence the statistical factor is one. Also $m_H\simeq m_p$, so ${\lambda }_H^3/{\lambda }_p^3\simeq 1$. Charge neutrality gives $n_p=n_e$, and thereforeThis is the specified ground-state Saha equation. Keeping excited atomic partition functions or a different spin-counting convention can change a prefactor, so its convention must be stated rather than silently adding a factor of two. The left side has dimensions of a number density, not a dimensionless electron-to-neutral ratio.

Let $N=n_H+n_p$ be the conserved number density of hydrogen nuclei, whether ionized or bound, and $x=n_p/N=n_e/N$ the ionization fraction. The ideal gas pressure counts neutral atoms, free protons and free electrons:At low temperature $x\simeq 0$ and $P\simeq N{k}_{B}T$. Heating through the ionization region increases the particle count as well as their thermal energy, giving a steeper rise. At high temperature $x\simeq 1$, so $P\simeq 2N{k}_{B}T$ and the asymptotic slope is twice the low-temperature slope. There is a smooth ionization crossover, not a pressure discontinuity.

The held graph is legible. Representative readings are $P(5000\mathrm{K})\simeq 6.8\mathrm{Pa}$ and $P(15000\mathrm{K})\simeq 41.4\mathrm{Pa}$, with readout uncertainties of roughly $0.3$–$0.5\mathrm{Pa}$ and about $100\mathrm{K}$. They give $N\simeq P/(k_{B}T)$ on the neutral branch and $N\simeq P/(2k_{B}T)$ on the ionized branch, respectively. Both are consistent withThe error is a conservative graph-reading estimate, not a fitted experimental confidence interval.

For that density, the Saha equation gives $x\simeq 0.50$ at $9000\mathrm{K}$, $x\simeq 0.81$ at $10000\mathrm{K}$, and $x\simeq 0.9992$ at $15000\mathrm{K}$. Differentiating the ionization relation givesThe calculated maximum slope occurs near $9.25\times 10^3\mathrm{K}$, consistent with the source's visibly steepest region at roughly $(9.3\pm 0.4)\times 10^3\mathrm{K}$. This checks the actual slope rather than identifying it without justification with exactly half ionization.

Taking the neutral ground-state gas as the zero of internal binding energy, the matter energy density is translational kinetic energy plus ionization energy:At $15000\mathrm{K}$, with $R=13.6\mathrm{eV}$, these terms are about $62$ and $218\mathrm{J}{\mathrm{m}}^{-3}$, givingRest energies are excluded. A separate equilibrium radiation field, if the gas were in a blackbody cavity, would add its own radiation energy density; it is not part of this gas-only estimate.

Increasing the conserved nucleus density to $10N$ multiplies the low-temperature neutral-gas slope and high-temperature ionized-gas slope by ten. However, at a fixed intermediate temperature, $x^2/(1-x)=S(T)/(10N)$ is smaller, so a smaller fraction is ionized. The curved transition therefore moves to higher temperature; the new graph is not simply ten times the old graph throughout the crossover.

At a fixed chosen ionization fraction, $S(T)$ must increase tenfold. Locally this requireswhich is of order $10^3\mathrm{K}$ in this region. Directly evaluating the same Saha equation at $10^{21}{\mathrm{m}}^{-3}$ moves the maximum pressure slope from about $9250$ to $10560\mathrm{K}$. The two asymptotic slopes still have ratio two, because each hydrogen nucleus still contributes one particle when neutral and two when fully ionized.