 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-2/section-c/c10/b/solution

Impose periodic boundary conditions on a large cubic box. The allowed wavevectors occupy a lattice of cell volume

(2 π)^{3} / V

in wavevector space. Including the two spin states, the number of states with magnitude at most

k

N (k) = 2 \frac{V}{( 2 π ) ^{3}} \frac{4 π k ^{3}}{3} = \frac{V k ^{3}}{3 π ^{2}} .

Using

k = 2 m_{e} ϵ /ℏ

g (ϵ) = \frac{V}{2 π ^{2}} (\frac{2 m _{e}}{ℏ ^{2}})^{3/2} ϵ, ϵ \geq 0.

It counts states per unit energy; divide by

V

for states per unit volume per unit energy. At zero temperature, the Fermi-Dirac distribution fills all states up to the Fermi energy, giving

N = \int_{0}^{E_{F}} g (ϵ) d ϵ = \frac{V}{3 π ^{2}} (\frac{2 m _{e} E _{F}}{ℏ ^{2}})^{3/2} .

n = N / V

gives

k_{F} = (3 π^{2} n)^{1/3}, E_{F} = \frac{ℏ ^{2}}{2 m _{e}} (3 π^{2} n)^{2/3}, g (E_{F}) = \frac{3 N}{2 E _{F}} .

The energy zero is the bottom of the free-electron band. The electron Fermi temperature is

T_{F} = E_{F} / k_{B}

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