 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/30a/b/solution

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

Write

r = e^{x}

ϵ = ℏ/ 2 m

L = l + 1/2

, and

λ = - ∣ λ ∣

. Since

V (r) = - V_{0} / r

, the transformed equation has

q (x) = ∣ λ ∣ e^{2 x} - V_{0} e^{x} + ϵ^{2} L^{2} .

The two turning radii

a < b

∣ λ ∣ r^{2} - V_{0} r + ϵ^{2} L^{2} = ∣ λ ∣ (r - a) (r - b),

a + b = \frac{V _{0}}{∣ λ ∣}, ab = \frac{ϵ ^{2} L ^{2}}{∣ λ ∣} .

Inside the allowed interval,

∣ q ∣ = ∣ λ ∣ (r - a) (b - r), d x = \frac{d r}{r} .

\frac{1}{ϵ} \int_{x_{a}}^{x_{b}} ∣ q (x) ∣^{1/2} d x = \frac{∣ λ ∣}{ϵ} \int_{a}^{b} \frac{( r - a ) ( b - r )}{r} d r = \frac{π ∣ λ ∣}{2 ϵ} (b - a)^{2} = π (\frac{V _{0}}{2 ϵ ∣ λ ∣} - L) .

Equating this with

(n + 1/2) π

from part (a), and using

L = l + 1/2

, gives

\frac{V _{0}}{2 ϵ ∣ λ ∣} = n + l + 1.

Since

ϵ^{2} = ℏ^{2} / (2 m)

∣ λ ∣ = \frac{V _{0}^{2}}{4 ϵ ^{2} ( n + l + 1 ) ^{2}} = \frac{m V _{0}^{2}}{2 ℏ ^{2} ( n + l + 1 ) ^{2}} .

E = λ = - \frac{m V _{0}^{2}}{2 ℏ ^{2} ( n + l + 1 ) ^{2}} .

 New to topics? Read the docs here!