 Solution

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

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

P (r < r_{p}) = ∣ A ∣^{2} \int_{0}^{r_{p}} r^{2 l + 2} e^{- 2 r / [a (l + 1)]} d r .

Throughout this small interval the exponential is

1 + O (r_{p} / a)

for fixed

l

. Hence

P (r < r_{p}) = \frac{∣ A ∣ ^{2} r _{p}^{2 l + 3}}{2 l + 3} [1 + O (r_{p} / a)] .

Using the normalization above,

P (r < r_{p}) = \frac{2 ^{2 l + 3}}{( 2 l + 3 )! ( l + 1 ) ^{2 l + 3}} (\frac{r _{p}}{a})^{2 l + 3} [1 + O (r_{p} / a)], C = (2 l + 3)! (l + 1)^{2 l + 3} .

The high power of the small radius reflects the near-origin behavior

R \propto r^{l}

of the hydrogen atom state.

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