In the stated units, $\alpha=1/r$. The <change of variables formula> gives
$$
p(\alpha)=p(1/\alpha)\left|\frac{d(1/\alpha)}{d\alpha}\right|
=\frac1{2r_0^3}\alpha^{-4}e^{-1/(r_0\alpha)}.
$$
Thus, with $\alpha_0=1/r_0$,
$$
\boxed{p(\alpha)=C\left(\frac{\alpha}{\alpha_0}\right)^\gamma e^{-\alpha_0/\alpha}},
\qquad
C=\frac1{2\alpha_0},\quad\gamma=-4.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 219 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-219/1/b/solution