= Solution
The time-independent <Schrödinger equation> is $[-\hbar^2\nabla^2/(2m)+V]\psi=E\psi$, with attractive <Coulomb potential> $V(r)=-\kappa/r$, where $\kappa=e^2/(4\pi\varepsilon_0)$. Separating $\psi=R(r)Y_\ell^m(\theta,\varphi)$ uses the <spherical harmonic> angular <eigenvalue> $-\ell(\ell+1)$ of the unit-sphere <Laplacian>. The radial kinetic contribution is $-\hbar^2(R''+2R'/r)/(2m)$, the angular kinetic contribution is the centrifugal term $\hbar^2\ell(\ell+1)R/(2mr^2)$, and the remaining term is the attractive <potential energy>. The right side is the <energy eigenvalue> times the radial <wavefunction>. Here the stated $m$ neglects nuclear recoil; including it replaces the electron mass by the <reduced mass>.
For $\ell=0$, substitution of the ground-state exponential gives
$$
-\frac{\hbar^2}{2m}(\alpha^2-2\alpha/r)-\frac\kappa r=E_1.
$$
Its $1/r$ coefficient must vanish and its constant coefficient determines the <energy>, so
$$
\boxed{\alpha=\frac{m\kappa}{\hbar^2}=\frac1{a_0},\qquad E_1=-\frac{m\kappa^2}{2\hbar^2}.}
$$
Here $a_0=\hbar^2/(m\kappa)$ is the <Bohr radius>.
For $R_2=N_2(r+b)e^{-\beta r}$, direct differentiation yields
$$
R_2''+\frac2rR_2'=N_2e^{-\beta r}\left[\beta^2(r+b)-4\beta+\frac{2-2\beta b}{r}\right].
$$
Cancel the nonzero exponential in the radial <Schrödinger equation> and compare coefficients of $r$, the constant term, and $1/r$. They give, respectively,
$$
E_2=-\frac{\hbar^2\beta^2}{2m},\qquad \frac{2\hbar^2\beta}{m}=\kappa,\qquad \frac{\hbar^2}{m}(\beta b-1)-\kappa b=0.
$$
Consequently
$$
\boxed{\beta=\frac1{2a_0},\qquad b=-2a_0,\qquad E_2=-\frac{m\kappa^2}{8\hbar^2}=\frac14E_1.}
$$
The positive decay constants ensure normalizability; the second radial <wavefunction> has one positive-radius node at $2a_0$, as appropriate to the first radial excitation.
If the entire energy gap is carried by one <photon>, neglecting atomic recoil, <conservation of energy> gives the requested frequency
$$
\boxed{\nu_{\rm gap}=\frac{E_2-E_1}{h}=\frac{3m\kappa^2}{8h\hbar^2}=\frac{3me^4}{8h\hbar^2(4\pi\varepsilon_0)^2}.}
$$
There is a physical qualification: these are the $2s$ and $1s$ states, so the usual single-photon electric-dipole transition is forbidden by the <electric-dipole selection rules for hydrogen>. Explicitly, both angular factors are $Y_0^0$, and $\int |Y_0^0|^2\hat{\mathbf r}\,d\Omega=0$, so every component of $\langle1s|\mathbf r|2s\rangle$ vanishes. The leading isolated-atom decay is instead two-photon emission, for which $\nu_1+\nu_2=\nu_{\rm gap}$ and individual frequencies are not fixed. This metastability and the two-photon channel are documented in https://tsapps.nist.gov/publication/get_pdf.cfm?pub_id=842564[NIST's hydrogen spectral-data compilation]. Thus the boxed answer is the energy-gap frequency under the printed single-photon assumption, rather than an assertion of an allowed electric-dipole line.
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