= Solution
The <Time-independent Schrödinger equation> is $[-\hbar^2\nabla^2/(2m)+V]\Psi=E\Psi$. For the <hydrogen atom>, separate variables as $\Psi=g(r)Y_{\ell m}(\theta,\varphi)$. The spherical-coordinate <Laplacian> has radial part $r^{-2}\partial_r(r^2\partial_r)$ and angular part $r^{-2}\Delta_{S^2}$. A <spherical harmonic> obeys $\Delta_{S^2}Y_{\ell m}=-\ell(\ell+1)Y_{\ell m}$. The first displayed term is therefore radial kinetic energy; the second is the attractive <Coulomb's law> potential energy $-e^2/(4\pi\varepsilon_0r)$ of electron and proton; the third is the angular kinetic energy, also called the <centrifugal potential>. Regular single-valued angular states have $\boxed{\ell=0,1,2,\ldots}$, with <orbital angular momentum> squared $\hbar^2\ell(\ell+1)$.
For $g=r^\alpha e^{-\beta r}$, direct differentiation gives
$$
\frac{g''+2g'/r}{g}=\frac{\alpha(\alpha+1)}{r^2}-\frac{2\beta(\alpha+1)}r+\beta^2.
$$
Substitution into the <Radial Schrodinger equation for the hydrogen atom> and equality of the coefficients of $r^{-2}$, $r^{-1}$ and the constant term give
$$
\alpha(\alpha+1)=\ell(\ell+1),\qquad \frac{\hbar^2\beta(\alpha+1)}m=\frac{e^2}{4\pi\varepsilon_0},\qquad E=-\frac{\hbar^2\beta^2}{2m}.
$$
The regular radial solution selects $\alpha=\ell$, rather than the singular branch $-\ell-1$. At $\ell=0$, $r^{-1}$ is locally square integrable but is still inadmissible as a regular Coulomb eigenfunction: its <Laplacian> has a point-source contribution at the origin. Bound-state decay requires $\beta>0$. Thus, with the <Bohr radius> $a_0=4\pi\varepsilon_0\hbar^2/(me^2)$,
$$
\boxed{\alpha=\ell,\qquad\beta=\frac1{a_0(\ell+1)},\qquad E_\ell=-\frac{me^4}{2(4\pi\varepsilon_0)^2\hbar^2(\ell+1)^2}.}
$$
These are the nodeless states with <principal quantum number> $n=\ell+1$. Increasing $\ell$ raises the energy towards zero, so emission is from the state labelled $\ell+1$ to the one labelled $\ell$. Using the <Planck constant> $h=2\pi\hbar$ and <photon> energy $h\nu$, the frequency is
$$
\boxed{\nu=\frac{me^4}{2(4\pi\varepsilon_0)^2\hbar^2h}\left[\frac1{(\ell+1)^2}-\frac1{(\ell+2)^2}\right].}
$$
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