 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2021/paper-2/a9/a/iii/solution

Let

A

be the total two-dimensional area. There are

A / (2 π)^{2}

wavevector states per reciprocal-space area per spin. Including spin degeneracy two and filling a disk gives

N_{e} = 2 \frac{A}{( 2 π ) ^{2}} π k_{F}^{2} = \frac{A k _{F}^{2}}{2 π} .

n_{e} = N_{e} / A

this yields

k_{F} = 2 π n_{e}, E_{F} = \frac{π ℏ ^{2} n _{e}}{m _{e}} .

The disk is centred at zero wavevector, and its boundary is a fermi circle. Omitting the two spin states would incorrectly double the energy inferred for a given number density.

 New to topics? Read the docs here!