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ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2021/paper-2/section-b/20/ii/solution

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

Codex  0 2026-10-20

Use the magnitudes of the single-electron-band Drude Hall relation and longitudinal sheet resistivity:

∣ ρ_{x y} ∣ = \frac{B}{n _{s} e}, ρ_{xx} = \frac{1}{n _{s} e μ} .

(1)

The given values are two-dimensional resistivities, so no sample thickness is inserted. Thus

n_{s} = \frac{1}{( 1.602 \times 1 0 ^{- 19} ) ( 2083 )} = 2.997 \times 1 0^{15} m^{- 2} = 2.997 \times 1 0^{11} c m^{- 2},

(2)

μ = \frac{∣ ρ _{x y} ∣}{B ρ _{xx}} = \frac{2083}{0.6} = 3.472 \times 1 0^{3} m^{2} V^{- 1} s^{- 1} = 3.472 \times 1 0^{7} c m^{2} V^{- 1} s^{- 1} .

(3)

The carrier sign follows from a specified Hall-voltage orientation; these calculations use positive magnitudes. A parallel conducting layer, several occupied subbands or a quantized Hall plateau would require a different interpretation of a single-field measurement.

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