Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 2 / A10 / b

Near the zone centre, $\cos (k_{i}a)\simeq 1-k_i^2{a}^2/2$. The tight-binding model becomesOnly $|\mathbf{k}{|}^2$ appears at quadratic order, so the band effective mass tensor is isotropic. Higher-order lattice terms need not be isotropic.

For the conduction band, ${\beta }_c<0$, so the curvature is positive andHere $a=5.42\times 10^{-10}$ m and both beta parameters must be converted from eV to joules. The band edges at the zone centre areso the model has a direct gap $\boxed{E_g=1.240\mathrm{eV}}$. The mass values describe curvature at these extrema; they are not constant masses throughout the full tight-binding bands.

For the valence band, ${\beta }_v>0$, so the electron curvature is negative. A missing electron is a positive-mass electron hole:The quadratic expansion has $E_v(\mathbf{k})\simeq E_v(0)-{\beta }_v{a}^2|\mathbf{k}{|}^2$, so each diagonal electron inverse mass is $-2{\beta }_v{a}^2/{\hslash }^2$. Reversing the sign to describe the missing electron gives the positive hole mass above. As in the conduction calculation, convert beta from eV to joules and use $a=5.42\times 10^{-10}$ m; the result applies near the band edge.