 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/36a/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

T_{B}

, the excited states are just saturated and the condensate is about to form, so

N = \frac{M}{e ^{ϵ / (k T_{B})} - 1} .

Therefore

T_{B} = \frac{ϵ}{k lo g ( 1 + M / N )} .

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