I have a question regarding the final sentence written in the solution to part iii) (found below) for the question given below.
In the Bohr model of the hydrogen atom, the radius of the electron orbit in state $n$ is equal to $n^2a_0$, where $a_0$ is the Bohr radius. The speed of the electron in state $n$ is $\alpha c/n$, where $\alpha$ is the fine structure constant. The energy of state $n$ is equal to $−{R_{\infty}}hc/n^2$, where $R_{\infty}$ is the Rydberg constant. Expressions for $a_0$, $\alpha$ and $R_{\infty}$ are given below. Using these results, calculate the potential energy, the kinetic energy and the total energy of the $n=2$ state in hydrogen in terms of the Rydberg constant. Comment on the relation between these quantities.
Useful information: $$a_0=\frac{4\pi\epsilon_0{\hbar}^2}{me^2}$$ $$\alpha=\frac{e^2}{4\pi\epsilon_0\hbar\,c}$$ $$R_{\infty}=\frac{me^4}{2(4\pi\epsilon_0)^2\hbar^2}\frac{1}{hc}$$ where the symbols have their usual meanings.
Solution to part 1] i):
$$r_n=n^2\,a_0$$ so $${\mathrm{PE}}=\frac{-e^2}{4\pi\epsilon_0\,r_n}=\frac{-e^2}{4\pi\epsilon_0\,n^2\,a_0}=-\frac{e^2me^2}{n^2(4\pi\epsilon_0)^2\hbar^2}$$ Since $n=2$, $$\bbox[5px,border:2px solid red] {\mathrm{PE}=-\frac{R_{\infty}hc}{2}}\tag{1}$$ Now n=2 for $$\mathrm{KE}=\frac12m{v_n}^2=\frac{m\,\alpha^2\,c^2}{2n^2}=\frac{R_{\infty}hc}{4}$$ and $E=-\frac{R_{\infty}hc}{4}$
So $\mathrm{E=PE+KE}$ as expected.
Question 1] iii)
Given the (normalised) expression for the $2$p radial wavefunction below, calculate the expectation value of $1/r$ in the $2$p state. $\bbox[yellow]{\text{ Compare your answer with the results in part (i) from the Bohr model.}}$ $$R_{2\mathrm{p}}(r)=\left(\frac{1}{a_0}\right)^{3/2}\frac{1}{2\sqrt{6}}\frac{r}{a_0}e^{-r/2a_0}$$ You may make use of the standard integral: $$\int_{0}^{\infty}x^n\exp(-bx)dx=\frac{n!}{b^{n+1}}$$
Solution to part 1] iii):
$$R_{2\mathrm{p}}(r)=\left(\frac{1}{a_0}\right)^{3/2}\frac{1}{2\sqrt{6}}\frac{r}{a_0}e^{-r/2a_0}$$ $$\begin{align}\Biggl\langle\frac{1}{r}\Biggr\rangle &=\int_0^\infty\lvert R_{2\mathrm{p}}(r)\rvert^2\frac{1}{r}r^2dr\\&=\int_0^\infty\frac{1}{{a_0}^3}\frac{1}{24}\frac{r^2}{{a_0}^2}\,r\,e^{-r/a_0}dr\\&=\frac{1}{24{a_0}^3}6{a_0}^4\\&=\frac14a_0\end{align}$$
$\bbox[5px,border:2px solid red]{\text{Gives the same PE as the Bohr model as expected, since}\, r_{n=2}=n^2a_0=4a_0}$
I have marked in red the parts for which I don't understand. Why does it give the same PE as equation $(1)$? Before asking this question I did some digging around and found out that in atomic units $R_\infty hc=1/2$ and I note that from i) that $R_{\infty}hc$ has units of energy. Even with this, I still have no idea what the author saying in that last boxed sentence. Does it seem that $\frac14a_0=-\frac{R_{\infty}hc}{2}$?
Put simply, how does $\frac14a_0$ give a potential energy of $-\frac{R_{\infty}hc}{2}$?
