Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S16 / 1

Assume a dilute, nondegenerate, nonrelativistic ideal gas in local thermodynamic equilibrium, with ground-state atomic hydrogen, protons and electrons, and neglect excited atomic levels and interactions. Let the conserved number of hydrogen nuclei per volume be $n_0$, so $n_e=n_p=\zeta n_0$ and $n_H=(1-\zeta )n_0$. Equality of chemical potentials for $H\rightleftharpoons p+e$ gives the Saha ionization equationThe factor two is electron spin; $g_p$ and $g_H$ must be defined consistently with nuclear spin. The small difference between the atomic and proton translational partition functions has been neglected. ThereforeFor the ground-state convention $g_p=1$, $g_H=2$, the statistical prefactor $2g_p/g_H$ is one. At fixed volume $n_0$ is constant, whereas an isobaric heating process would not have a constant $A$.

One mole of original atoms contains $(1-\zeta )$ moles of neutral atoms, $\zeta$ moles of protons and $\zeta$ moles of electrons. All are monatomic translational species, so their combined thermal energy is $\frac{3}{2}RT(1+\zeta )$. Taking the neutral ground state as zero internal energy, ionization stores $\zeta N_A{E}_I$. ThusRadiation, excited states and nonideal interactions have been excluded consistently with the elementary Saha ionization equation.

At fixed volume, logarithmic differentiation of the Saha ionization equation givesDifferentiation the internal energy then yieldsAt low temperature $\zeta \to 0$ and $C_V\to 3R/2$. At high nonrelativistic temperature the gas is fully ionized, $\zeta \to 1$, and $C_V\to 3R$. The large intermediate excess reflects energy spent increasing ionization, as well as heating and creating additional translational particles. “Very high” here remains below regimes where relativistic, radiative or pair-creation contributions invalidate the model.

Put $x=E_I/(k_{B}T)$. With $\zeta =1/2$, the heat capacity formula becomesTaking $E_I=13.6\mathrm{eV}$ gives $\boxed{T=1.02\times 10^4\mathrm{K}}$. At half ionization, $n_e{n}_p/n_H=n_e$, so the Saha ionization equation directly givesThis uses $2g_p/g_H=1$ as specified in part (a); a different internal-state convention must be carried consistently into the numerical electron density. The original nuclei density is $n_0=2n_e\simeq 10^{21}{\mathrm{m}}^{-3}$.