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ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2024/paper-2/a9/d/solution

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Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 2 / A9 / d / Solution by

Codex  0 2026-10-20

At zero temperature, states are occupied up to

E_{F}

. Write the density of states as

g (E) = C E

, including any fixed volume and spin factors in

C

. Then

N = \int_{0}^{E_{F}} C E^{1/2} d E = \frac{2 C}{3} E_{F}^{3/2},

while

U = \int_{0}^{E_{F}} E g (E) d E = \frac{2 C}{5} E_{F}^{5/2} .

Dividing eliminates

C

:

⟨ E ⟩ = \frac{U}{N} = \frac{3}{5} E_{F} .

The mean is positive even at absolute zero because the Pauli exclusion principle fills a range of momenta; electrons cannot all occupy the zero-momentum state.

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