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ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2022/paper-2/section-a-ii/16/iii/solution

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Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2022 / Paper 2 / Section A(ii) / 16 / iii / Solution by

Codex  0 2026-10-20

Interpret the stated Fermi energy as measured above the bottom

E_{0}

of the occupied subband, and take a degenerate, nearly zero-temperature gas. Then

n_{s} = \int_{E_{0}}^{E_{F}} D_{2 D} (E) d E = \frac{m ^{*}}{π ℏ ^{2}} (E_{F} - E_{0}) .

(1)

For a

0.100 eV

filling energy,

n_{s} = (2.80 \times 1 0^{13}) (0.100) = 2.80 \times 1 0^{12} c m^{- 2} = 2.80 \times 1 0^{16} m^{- 2} .

(2)

A Fermi energy referenced instead to an unspecified bulk conduction-band edge would not determine this number without the subband offset. The assumption that only one subband is occupied is imposed by the question; at this comparatively large density, an actual confinement potential should be checked before applying it to a device.

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