Chapter 36, Problem 42P

To determine

To show: The substitution of $\psi \left(r\right)=A{e}^{\frac{-r}{a_0}}$ in equation $-\frac{{\hslash }^2}{2m{r}^2}\frac{d}{dr}\left(r^2\frac{d\psi }{dr}\right)-\frac{k{e}^2}{r}\psi =E\psi$ will results $E=-\frac{{\hslash }^2}{2m{a}_0^2}+\frac{{\hslash }^2}{mr{a}_0}-\frac{k{e}^2}{r}$.