 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-3/34c/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

R_{0} (r) = u (r) / r

. In the attractive shell

a < r \leq 2 a

u (a) = 0

gives

u_{in} (r) = A sin [κ (r - a)], κ^{2} = k^{2} + \frac{2 m V _{0}}{ℏ ^{2}} .

Outside the potential, choose

u_{out} (r) = B sin (k r + δ_{0}) .

r = 2 a

yields

κ cot (κa) = k cot (2 ka + δ_{0}) .

Put

C = κ cot (κa)

tan (2 ka + δ_{0}) = \frac{k}{C},

and solving for

tan δ_{0}

gives

tan δ_{0} = \frac{k - C t a n ( 2 ka )}{C + k t a n ( 2 ka )} = \frac{k cos ( 2 ka ) - κ cot ( κa ) sin ( 2 ka )}{k sin ( 2 ka ) + κ cot ( κa ) cos ( 2 ka )} .

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