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.

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.

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.

At fixed temperature,Thus magnetic-field-dependent Langmuir adsorption is even in $B$ and increases with $|B|$. At fixed nonzero gas pressure, stronger field increases surface coverage; in an isotherm plot, the half-coverage pressure $p_{1/2}=1/b$ moves to lower pressure. The saturation coverage remains one because the number of sites is unchanged.

For weak fields, with ${\theta }_0=b(0)p/[1+b(0)p]$,There is no linear small-field change. At large $|x|$, $\cosh x\sim e^{|x|}/2$, and at fixed $p>0$ the coverage approaches one. The field lowers the free energy of the surface spin ensemble relative to the unsplit bulk; simply averaging the two surface energies to $-E_b$ would miss this equilibrium effect.

At adsorption equilibrium, ${\mu }_{\mathrm{ads}}={\mu }_g$. The expressions from part (d) giveDefineSolving the occupation ratio yieldsThe coefficient $b$ has units of inverse pressure, so $bp$ is dimensionless. The two spin states contribute a factor two on both sides of the adsorption equilibrium and cancel; retaining the surface factor but dropping the bulk degeneracy would incorrectly double $b$.

For the adsorbed lattice gas, the Helmholtz free energy isUse Stirling's approximation with $M,N,M-N$ large, then differentiate at fixed $M,T,B$. Writing surface coverage as $\theta =N/M$ gives the adsorption chemical potentialThe occupation ratio comes from site-mixing entropy; it is essential near saturation.

Let $N_g$ be the number of free atoms and $n_g=N_g/V$. In the field-free gas, both spin states have zero energy, so the one-atom partition function is $q_g=2V/{\Lambda }^3$, where $\Lambda$ is the thermal de Broglie wavelength. For a dilute ideal gas,ThereforeThe last equality uses $p=n_g{k}_{\mathrm{B}}T$. The bulk spin degeneracy is two even though the bulk magnetic field vanishes: degeneracy is not the absence of spin states.

Use the ideal Langmuir adsorption model: identical independent sites, at most one atom per site, no lateral interactions, and no additional thermally active adsorbed degrees of freedom beyond the two spin states. Treat the atoms as indistinguishable and localized after adsorption; only site occupancy and spin choices distinguish configurations.

There are $\left(\genfrac{}{}{0ex}{}{M}{N}\right)$ ways to select $N$ occupied sites. For any such selection, independent spin sums contribute $z^N$. ThusWith $q_{\mathrm{ads}}=Mz$ as defined in part (b), this isEquivalently, starting from labelled particles gives $M!/(M-N)!$ distinct assignments to different sites; division by $N!$ removes particle relabellings. The $M^{-N}$ removes the site multiplicity already included in each full one-particle function.

If $q_{\mathrm{ads}}$ denotes the one-site internal factor instead, $f(M,N)=\left(\genfrac{}{}{0ex}{}{M}{N}\right)$ and $Q_N=\left(\genfrac{}{}{0ex}{}{M}{N}\right)q_{\mathrm{ads}}^N$. These are identical physical results, not different counting models. In particular, the complete-function convention gives $f(M,1)=1$ and hence $Q_1=q_{\mathrm{ads}}$.