= Solution
Comparing the prepared energy with the formula in part (ii),
$$
\frac1{8(l+1)^2}=\frac1{72},
$$
so $l+1=3$ and $l=2$. The radial probability density, apart from normalization, is
$$
r^2|R(r)|^2=r^{2l+2}e^{-2r/[a(l+1)]}.
$$
Using the given factorial <integral>, its mean is the ratio
$$
\langle r\rangle
=\frac{\int_0^\infty r^{2l+3}e^{-2r/[a(l+1)]}\,dr}
{\int_0^\infty r^{2l+2}e^{-2r/[a(l+1)]}\,dr}
=\frac{(2l+3)(l+1)}2a.
$$
For $l=2$ this gives the <mean radius of a circular Coulomb bound state>
$$
\boxed{\langle r\rangle=\frac{21}{2}a},
\qquad
\boxed{t=\frac{21}{2}}.
$$
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