= Solution
Use the nonrelativistic <Coulomb potential energy> $V=-Ze^2/(4\pi\epsilon_0r)$, with the nucleus treated as fixed. Since the <wavefunction> has zero <orbital angular momentum>, its <Laplacian> is
$$
\nabla^2\psi=\frac1{r^2}\frac{d}{dr}\left(r^2\frac{d\psi}{dr}\right)=\left(\beta^2-\frac{2\beta}{r}\right)\psi.
$$
Substituting into the time-independent <Schrödinger equation> gives
$$
\frac{H\psi}{\psi}=-\frac{\hbar^2\beta^2}{2m_e}+\frac1r\left(\frac{\hbar^2\beta}{m_e}-\frac{Ze^2}{4\pi\epsilon_0}\right).
$$
For a constant <energy eigenvalue>, the coefficient of $1/r$ must vanish. Therefore $\boxed{\beta=Z/a_0}$, where $a_0$ is the <Bohr radius>. <Wavefunction normalization> then requires
$$
1=4\pi|\alpha|^2\int_0^\infty r^2e^{-2\beta r}\,dr=\frac{\pi|\alpha|^2}{\beta^3}.
$$
Choosing the irrelevant overall phase so that $\alpha$ is real and positive gives $\boxed{\alpha^2=\beta^3/\pi}$. The remaining constant term is the <hydrogen ground-state energy>
$$
\boxed{E=-\frac{\hbar^2\beta^2}{2m_e}=-\frac{m_eZ^2e^4}{2(4\pi\epsilon_0)^2\hbar^2}\simeq-13.6Z^2\,\mathrm{eV}.}
$$
Replacing $m_e$ by the electron-nucleus <reduced mass> gives the usual finite-nuclear-mass refinement.
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