Past exam of the physics A course of the University of Cambridge / 2022 / Paper 2 / Section C / C8 / b / ii

Solution

k

(L /2 π)^{2}

k

and thickness

d k

2 πk d k

. Therefore

g_{2}^{(1)} (ϵ) = (\frac{L}{2 π})^{2} 2 πk \frac{d k}{d ϵ} = \frac{L ^{2} m}{2 π ℏ ^{2}} .

g_{2}^{(1)} \propto ϵ^{0}

, above the band bottom. Including both spin states gives

g_{2} = L^{2} m / (π ℏ^{2})

. Dividing by

L^{2}

gives the density per unit physical area, rather than the total specimen's density.

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