Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2021 / Paper 2 / Section B / 20 / iii / Solution

For a parabolic two-dimensional electron gas with one valley and two spin states, the density of states per area is $g_{2D}=m^{\ast }/(\pi {\hslash }^2)$. Integrating from the subband edge to the zero-temperature Fermi energy gives $n_s=g_{2D}{E}_F$, henceHere $m^{\ast }=0.067(9.109\times 10^{-31})=6.103\times 10^{-32}\mathrm{kg}$, and the energy is measured from the occupied subband bottom. Equivalently $k_F=\sqrt{2\pi n_s}=1.3722\times 10^8{\mathrm{m}}^{-1}$. ThereforeThe spin factor is already included in $g_{2D}$; counting it again would halve the energy incorrectly.