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Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 14A / iii / Solution

Comparing the prepared energy with the formula in part (ii),

\frac{1}{8 ( l + 1 ) ^{2}} = \frac{1}{72},

so

l + 1 = 3

and

l = 2

. The radial probability density, apart from normalization, is

r^{2} ∣ R (r) ∣^{2} = r^{2 l + 2} e^{- 2 r / [a (l + 1)]} .

Using the given factorial integral, its mean is the ratio

⟨ r ⟩ = \frac{\int _{0}^{\infty} r ^{2 l + 3} e ^{- 2 r / [a (l + 1)]} d r}{\int _{0}^{\infty} r ^{2 l + 2} e ^{- 2 r / [a (l + 1)]} d r} = \frac{( 2 l + 3 ) ( l + 1 )}{2} a .

For

l = 2

this gives the mean radius of a circular Coulomb bound state

⟨ r ⟩ = \frac{21}{2} a, t = \frac{21}{2} .

Solved by gpt-5.6-sol high.

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