The radial wavefunction $R(r)$ for an electron in a hydrogen atom satisfies the equation

$$-\frac{\hbar^{2}}{2 m r^{2}} \frac{d}{d r}\left(r^{2} \frac{d}{d r} R(r)\right)+\frac{\hbar^{2}}{2 m r^{2}} \ell(\ell+1) R(r)-\frac{e^{2}}{4 \pi \epsilon_{0} r} R(r)=E R(r)$$

Briefly explain the origin of each term in this equation.

The wavefunctions for the ground state and the first radially excited state, both with $\ell=0$, can be written as

$$\begin{aligned} &R_{1}(r)=N_{1} e^{-\alpha r} \\ &R_{2}(r)=N_{2}\left(1-\frac{1}{2} r \alpha\right) e^{-\frac{1}{2} \alpha r} \end{aligned}$$

where $N_{1}$ and $N_{2}$ are normalisation constants. Verify that $R_{1}(r)$ is a solution of $(*)$, determining $\alpha$ and finding the corresponding energy eigenvalue $E_{1}$. Assuming that $R_{2}(r)$ is a solution of $(*)$, compare coefficients of the dominant terms when $r$ is large to determine the corresponding energy eigenvalue $E_{2}$. [You do not need to find $N_{1}$ or $N_{2}$, nor show that $R_{2}$ is a solution of $\left.(*) .\right]$

A hydrogen atom makes a transition from the first radially excited state to the ground state, emitting a photon. What is the angular frequency of the emitted photon?