 Solution

The smallest energy eigenvalue in part (iii) occurs at the smallest possible principal quantum number, $N_{\mathrm{p}}=1$. Since $n,\ell \ge 0$, this forces $n=0$ and $\ell =0$. Therefore $f=a_{0}r$, andFor the spherically symmetric ground state, the full three-dimensional wavefunction normalization uses $d^{3}x=r^{2}drd\Omega$. Integrating over the solid angle givesPut $R=2{\alpha }_{0}r$ and use the supplied factorial integral:Thus $|a_0{|}^2={\alpha }_0^3/\pi$. Choosing the positive global phase convention givesThis is the radial normalization of the hydrogen ground state convention in which the angular factor is absorbed into the spherically symmetric function. If instead one writes $\Psi =R(r)Y_{00}$ with the normalized spherical harmonic $Y_{00}=1/\sqrt{4\pi }$, then $R(r)=2{\alpha }_0^{3/2}{e}^{-{\alpha }_{0}r}$. Both conventions describe the same normalized three-dimensional quantum state.