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Take the spherical harmonic $Y_{00}$ to have unit angular norm. The radial normalization condition is thenThereforeEquivalently, if $R$ denotes the entire spherically symmetric wavefunction rather than the radial factor multiplying normalized $Y_{00}$, then $C=({\gamma }^3/\pi {)}^{1/2}$.

The expected radius isFor the ground state, $\gamma =m{q}^2/{\hslash }^2$. The Bohr radius is $a_0={\hslash }^2/(m{q}^2)=1/\gamma$, soThe mean radius is therefore of the Bohr-radius scale and is one and a half times $a_0$. This is the radial normalization of the hydrogen ground state.

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.