 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/37a/d/solution

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

Put

x = β ϵ

z = e^{x} + 1 + e^{- x} = 1 + 2 cosh x,

and independence gives

Z = z^{N}

E = - \frac{\partial lo g Z}{\partial β} = - \frac{2 N ϵ sinh x}{1 + 2 cosh x}

and, using part (c),

S = N k_{B} [lo g (1 + 2 cosh x) - \frac{2 x sinh x}{1 + 2 cosh x}] .

T \to \infty

x \to 0

, all three levels become equally likely and

E \to 0, S \to N k_{B} lo g 3.

T \to 0^{+}

, every particle occupies the nondegenerate ground level

- ϵ

, and

E \to - N ϵ, S \to 0.

 New to topics? Read the docs here!