 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/36a/b/i/solution

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

h^{2}

Z_{1} = \frac{1}{h ^{2}} \int_{R^{2}} e^{- β p^{2} / (2 m)} d^{2} p \int_{r < R} e^{- β κ r^{2} /2} d^{2} x = \frac{1}{h ^{2}} (\frac{2 πm}{β}) (2 π \int_{0}^{R} r e^{- β κ r^{2} /2} d r) = \frac{4 π ^{2} m}{h ^{2} β ^{2} κ} (1 - e^{- β κ R^{2} /2}) .

Equivalently, using

h = 2 π ℏ

Z_{1} = \frac{m}{ℏ ^{2} β ^{2} κ} (1 - e^{- β κ R^{2} /2}) .

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