 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/35c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

E_{F}

are occupied. Writing

g = \frac{A m}{π ℏ ^{2}},

we find

N = \int_{0}^{E_{F}} g d E = g E_{F}

and

E_{tot} = \int_{0}^{E_{F}} E g d E = \frac{1}{2} g E_{F}^{2} = \frac{1}{2} N E_{F} .

Because

g = C A

with

C = m / (π ℏ^{2})

E_{F} = N / (C A)

and

E_{tot} = \frac{N ^{2}}{2 C A} .

p = - (\frac{\partial E _{tot}}{\partial A})_{N} = \frac{N ^{2}}{2 C A ^{2}} = \frac{N E _{F}}{2 A} .

 New to topics? Read the docs here!