Solution

Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 2 / 17A / ii / Solution

{title2=$P(r<r_p)\sim\frac{2^{2l+3}(r_p/a)^{2l+3}}{(2l+3)!(l+1)^{2l+3}}$}

For a normalized <maximal-angular-momentum hydrogen state>, the probability inside a small radius $r_p$ is the integral of $|R|^2r^2$. Since $R$ behaves as $r^l$ near zero, it equals $2^{2l+3}(r_p/a)^{2l+3}/[(2l+3)!(l+1)^{2l+3}]$ times $1+O(r_p/a)$. Increasing angular momentum strongly suppresses penetration toward the Coulomb centre.

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 Incoming links: Near-origin probability in a maximal-angular-momentum hydrogen state