*Condensed Matter in a Nutshell*and came across a pretty interesting nugget: \[ \varepsilon (k) = \frac{\hbar^2 k^2}{2 m^*} \sim k_B T \,. \] (It's important to draw a distinction between \( k \), which is the lattice's wavenumber, and \(k_B\), which is the Boltzmann constant.) Basically, the total energy of a semiconductor lattice is related to its wavenumber—and it approximately scales up to the thermal properties of the system.

So, provided that we know this, does that mean that the thermal voltage of a semiconductor would be more accurately described as

\[ V_T = \frac{\varepsilon (k)}{q_0} = \frac{\hbar^2 k^2}{2 m^* q_0} \,?\] The answer probably seems trivial to people that have spent more time reading about semiconductor theory than I have.