 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-3/35a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

V

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

ℏ ω = ℏ c k

(e^{ℏ c k / T} - 1)^{- 1}

. Omitting the temperature-independent zero-point energy, the total energy is therefore

E = \frac{V ℏ c}{π ^{2}} \int_{0}^{\infty} \frac{k ^{3}}{e ^{ℏ c k / T} - 1} d k .

x = ℏ c k / T

E = \frac{V T ^{4}}{π ^{2} ℏ ^{3} c ^{3}} \int_{0}^{\infty} \frac{x ^{3}}{e ^{x} - 1} d x .

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