 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ib/paper-2/17a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

- \frac{ℏ ^{2}}{2 m} (R^{''} + \frac{2}{r} R^{'} - \frac{l ( l + 1 )}{r ^{2}} R) - \frac{e ^{2}}{4 π ϵ _{0} r} R = ER .

Write

R = A r^{l} e^{- κ r}

. Then

\frac{R ^{''} + 2 R ^{'} / r - l ( l + 1 ) R / r ^{2}}{R} = κ^{2} - \frac{2 κ ( l + 1 )}{r} .

Substitution separates a constant term and a

1/ r

term. The latter vanishes precisely when

ℏ^{2} κ (l + 1) / m = e^{2} / (4 π ϵ_{0})

. With

κ = 1/ [a (l + 1)]

this determines the Bohr radius

a = \frac{4 π ϵ _{0} ℏ ^{2}}{m e ^{2}} .

The constant term determines

E = - \frac{ℏ ^{2}}{2 m a ^{2} ( l + 1 ) ^{2}} = - \frac{e ^{2}}{8 π ϵ _{0} a ( l + 1 ) ^{2}}, D = 8 (l + 1)^{2} .

n = l + 1

Because the angular integral is one, normalization requires

1 = ∣ A ∣^{2} \int_{0}^{\infty} r^{2 l + 2} e^{- 2 κ r} d r = ∣ A ∣^{2} \frac{( 2 l + 2 )!}{( 2 κ ) ^{2 l + 3}} .

Therefore

∣ A ∣^{2} = \frac{2 ^{2 l + 3}}{a ^{2 l + 3} ( 2 l + 2 )! ( l + 1 ) ^{2 l + 3}}, F = 2 l + 3, G = 2 l + 2, H = 2 l + 3.

The phase of

A

remains arbitrary.

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