 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/36b/c/solution

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

Here

k cot δ_{0} = - K

f_{0} (k) = \frac{1}{- K - ik} .

k = i K

, corresponding to the bound-state energy

E = - \frac{ℏ ^{2} K ^{2}}{2 m} .

Moreover,

σ_{T} ≃ 4 π ∣ f_{0} ∣^{2} = \frac{4 π}{K ^{2} + k ^{2}} .

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