 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/35a/b/iii/solution

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

ℏ ω

. The total internal energy is consequently

E_{tot} = \int_{0}^{ω_{D}} \frac{ℏ ω g ( ω )}{e ^{β ℏ ω} - 1} d ω .

x = β ℏ ω

and

β ℏ ω_{D} = T_{D} / T

, this becomes

E_{tot} = \frac{B ( k _{B} T ) ^{b + 2}}{ℏ ^{b + 1}} \int_{0}^{T_{D} / T} \frac{x ^{b + 1}}{e ^{x} - 1} d x, b = \frac{2}{α} - 1.

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