Set the ground electronic energy to zero and write the excitation energy as $\Delta =hc(120\mathrm{c}{\mathrm{m}}^{-1})$, or the characteristic temperature $\theta =\Delta /k_{\mathrm{B}}=172.65$ K. Counting each doubly degenerate manifold once gives the single-molecule electronic partition functionFor $N$ independent molecules the electronic factor is $q_{\mathrm{el}}^N$. Hence the electronic internal energy isThe equal degeneracies cancel from the excited-state probability. At low temperature the ground manifold dominates and $U_{\mathrm{el}}\to 0$; at high temperature both manifolds have equal populations, giving $U_{\mathrm{el}}\to N\Delta /2$. These are electronic contributions only, excluding translation, rotation and vibration.

Let ${\rho }_i=N_i/V$ be a molecular number density and define the volume-independent molecular partition function density $f_i=q_i/V$. Then $q_i/N_i=f_i/{\rho }_i$. Also define the reaction reference-energy differenceSubstitution into the equilibrium chemical-potential relation givesAn equilibrium constant expressed using concentration activities $a_i={\rho }_i/c^{\circ }$ is dimensionless:Here every $f_i$ has units ${\mathrm{m}}^{-3}$, and $c^{\circ }$ is the chosen standard concentration, initially one molecule per cubic metre. The factor $c^{\circ }$ is essential: without it the association quotient has units of volume.

Treat each species as a dilute classical ideal gas: molecules of the same species are indistinguishable, while different species are distinguishable. Let $N_i$ be the number of molecules of species $i$, $q_i(T,V)$ its molecular partition function with energies measured from a chosen internal reference, and ${\epsilon }^{0,i}$ that reference energy per molecule relative to a common chemical energy convention. Then the mixture's canonical partition function factorizes:The $N_i!$ corrects the overcounting of indistinguishable particles. The exponential restores the species-dependent reference energies that matter for reactions. Thus the Helmholtz free energy isUsing Stirling's approximation, $\ln (N_i!)\simeq N_i\ln N_i-N_i$, and differentiating at fixed $T,V,N_{j\ne i}$ yieldsThe ideal-gas $q_i$ does not depend on the number of molecules, so no additional $\partial q_i/\partial N_i$ term appears. These are chemical potentials per molecule; multiply by the Avogadro constant for a molar chemical potential. The result assumes the thermodynamic, nondegenerate-gas limit.

At cryogenic temperatures, the intrinsic carrier concentration is exponentially small, while a low phonon population permits relatively sharp band-edge and exciton spectral features. Near room temperature, thermal electron–hole generation increases strongly:This generally raises intrinsic conductivity, although increased phonon scattering tends to lower the individual carrier mobilities. With a gap of $1.24$ eV, the material remains far from a degenerate metal at room temperature.

For optical structure, compare $k_{\mathrm{B}}T\simeq 25.9$ meV at 300 K with the exciton binding energies. The $n=2$ binding of $9.56$ meV is substantially smaller, so excited bound states are readily thermally disrupted and broadened. The ground binding of $38.24$ meV is larger but only about $1.48k_{\mathrm{B}}T$: the ground exciton need not disappear completely, yet strong thermal ionization and broadening can weaken its sharp line. A comparison with $k_{\mathrm{B}}T$ is a useful scale argument, not an exact bound-pair fraction independent of density or lifetime.

Thermal lattice expansion and electron–phonon interactions can also shift a real material's gap and optical edge. The given fixed tight-binding parameters contain no temperature dependence, so they do not determine the magnitude or even a universal sign of that extra gap shift. Within the stated model, the dependable trends are more thermally generated carriers and less distinct excitonic structure on warming.

The Exciton Rydberg series lies below the unbound electron–hole threshold:Using $E_g=1.240$ eV and $R^{\ast }=0.038241$ eV givesTheir positive exciton binding energies relative to the continuum are $38.24$ meV and $9.56$ meV, respectively.

These are optical electron–hole excitation energies for negligible centre-of-mass momentum, not new occupied one-electron states inside the valence band. The simple model predicts their locations; observing a particular line also requires a nonzero optical transition matrix element and adequate linewidth.

Relative electron–hole motion uses the exciton reduced mass, rather than either band mass alone:In a screened hydrogenic Wannier-Mott exciton, replace $m_e$ by ${\mu }^{\ast }$ and ${\epsilon }_0$ by ${\epsilon }_r{\epsilon }_0$ in the hydrogen binding-energy scale. ThereforeThe effective Rydberg is an energy here, not a spectroscopic inverse length. Its scale is reduced both by dielectric screening and by the small band reduced mass. The associated radius is about ${\epsilon }_r(m_e/{\mu }^{\ast })a_0=23.5$ Å, several lattice spacings, consistent with using an extended-pair model as a first approximation rather than a tightly localized molecular exciton.

The smaller $|{\beta }_v|$ gives the valence edge a larger density-of-states prefactor than the conduction edge. If the Fermi level were exactly halfway across the gap, there would therefore be more thermally generated holes than electrons. Charge neutrality in an intrinsic semiconductor requires equal carrier densities, so the chemical potential moves upward: it suppresses holes while increasing electrons.

Explicitly, in the nondegenerate limit,and $n=p$ gives the intrinsic Fermi levelSince $N_v>N_c$, the Fermi level lies in the upper half of the gap at finite temperature.

The supplied inverse-$|\beta |$ density-of-states rule gives the correct ordering. Strictly, an isotropic three-dimensional parabolic band has a prefactor proportional to $(m^{\ast }{)}^{3/2}\propto |\beta {|}^{-3/2}$, rather than $|\beta {|}^{-1}$; this changes the size of the displacement but not its sign. At zero temperature the limiting intrinsic expression approaches midgap; a chemical potential in a completely empty gap is not uniquely fixed by carrier neutrality alone.