 Solution

ID: 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_{2 D} = m^{*} / (π ℏ^{2})

. Integrating from the subband edge to the zero-temperature Fermi energy gives

n_{s} = g_{2 D} E_{F}

, hence

E_{F} = \frac{π ℏ ^{2} n _{s}}{m ^{*}} = 1.716 \times 1 0^{- 21} J = 10.71 meV .

Here

m^{*} = 0.067 (9.109 \times 1 0^{- 31}) = 6.103 \times 1 0^{- 32} kg

, and the energy is measured from the occupied subband bottom. Equivalently

k_{F} = 2 π n_{s} = 1.3722 \times 1 0^{8} m^{- 1}

. Therefore

v_{F} = \frac{ℏ k _{F}}{m ^{*}} = \frac{2 E _{F}}{m ^{*}} = 2.371 \times 1 0^{5} m s^{- 1} .

The spin factor is already included in

g_{2 D}

; counting it again would halve the energy incorrectly.

 New to topics? Read the docs here!