= Solution
The smallest <energy eigenvalue> in part (iii) occurs at the smallest possible <principal quantum number>, $N_{\rm p}=1$. Since $n,\ell\geq0$, this forces $n=0$ and $\ell=0$. Therefore $f=a_0r$, and
$$
\boxed{\psi_0(r)=a_0e^{-\alpha_0r},\qquad \alpha_0=\frac{mKZ}{\hbar^2},\qquad E_0=-\frac{mK^2Z^2}{2\hbar^2}.}
$$
For the spherically symmetric <ground state>, the full three-dimensional <wavefunction normalization> uses $d^3x=r^2dr\,d\Omega$. Integrating over the <solid angle> gives
$$
1=4\pi|a_0|^2\int_0^\infty r^2e^{-2\alpha_0r}\,dr.
$$
Put $R=2\alpha_0r$ and use the supplied <factorial> integral:
$$
\int_0^\infty r^2e^{-2\alpha_0r}\,dr=\frac1{(2\alpha_0)^3}\int_0^\infty R^2e^{-R}\,dR=\frac{2!}{8\alpha_0^3}=\frac1{4\alpha_0^3}.
$$
Thus $|a_0|^2=\alpha_0^3/\pi$. Choosing the positive <global phase> convention gives
$$
\boxed{a_0=\left(\frac{\alpha_0^3}{\pi}\right)^{1/2}.}
$$
This 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 $R(r)=2\alpha_0^{3/2}e^{-\alpha_0r}$. Both conventions describe the same normalized three-dimensional <quantum state>.
Back to article page
