Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 2 / A7

Chemical equilibrium for $\mathrm{M}\rightleftharpoons {\mathrm{M}}^{+}+{\mathrm{e}}^{-}$ requiresInsert the given chemical potential form and define the positive ionization energyAfter collecting logarithms,soDefine a dimensionless concentration equilibrium constant using a standard number density $c^{\circ }$:Each $f_i$ is a molecular partition function density, with units of inverse volume. Consequently $f_e{f}_{{\mathrm{M}}^{+}}/f_{\mathrm{M}}$ has density units and division by $c^{\circ }$ makes $K_c$ dimensionless. All ground-state energies must use the same energy reference.

For monatomic species, there are no molecular rotational or vibrational factors. Assume excited electronic states are negligible, so the internal partition sums are the ground-state degeneracies. A free electron has two spin states, givingwith the analogous expression for the ion. Since losing one electron changes the atomic mass negligibly, $(m_{{\mathrm{M}}^{+}}/m_{\mathrm{M}}{)}^{3/2}\simeq 1$. HenceThis is the ground-state form of the Saha ionization equation. Its assumptions include a dilute, nondegenerate ideal mixture, negligible interparticle interactions, thermal and chemical equilibrium, no additional ionization stages, and consistent treatment of internal degeneracies. Nuclear-spin factors may be omitted when they cancel between the atom and ion; they must not be counted on only one side.

For fixed degeneracies and ionization energy,Therefore $K_c$ increases with temperature in this model. Ionization costs positive energy, so it is suppressed at low temperature by its Boltzmann factor. At higher temperature that penalty becomes less important.

Ionization also creates an additional independently translating particle and more ways to distribute the particles and their momenta. This favours the ionized mixture through its translational and mixing entropy, consistent with the $T^{3/2}$ prefactor and increasing ionization at high temperature.

Let $\alpha$ be the degree of ionization. Conservation of nuclei and charge neutrality, with no pre-existing electrons or other ions, giveSubstitution into the definition of $K_c$ givesThe physical root lies between zero and one:The second form avoids subtracting nearly equal terms at large $y$. The limits are $\alpha \simeq \sqrt{y}$ when $y\ll 1$ and $\alpha \simeq 1-1/y$ when $y\gg 1$. The dependence on initial number density matters: $K_c$ alone, without $c_{\mathrm{tot}}$ and the standard-state convention, does not specify the ionized fraction.

For hydrogen, the electronic ground state has $g_{\mathrm{H}}=2$, while the proton has no electronic degeneracy, $g_{{\mathrm{H}}^{+}}=1$. For sodium, the neutral $3s{}^2{S}_{1/2}$ ground state has $g_{\mathrm{Na}}=2$ and the closed-shell ion has $g_{\mathrm{N}{\mathrm{a}}^{+}}=1$. Ignoring hyperfine structure and consistently cancelling nuclear-spin factors, $2g_{{\mathrm{M}}^{+}}/g_{\mathrm{M}}=1$ for both.

At $5000\mathrm{K}$,It is convenient to calculate the density $D=K_c{c}^{\circ }$, which avoids choosing an unnecessary numerical value of the standard density. The results areThus hydrogen is about $0.00409\%$ ionized, whereas sodium is about $53.6\%$ ionized in the stated ground-state ideal-gas model. The large difference is dominated by the smaller sodium ionization energy in the exponential. The weak-ionization approximation is excellent for hydrogen but would be inaccurate for sodium, so the quadratic solution is needed there.