The binding energies of single ionized Helium are given by
$\begin{align*}-E_n = \frac{2^2 \mu e^4}{2 \hbar^2 n^2 (4 \pi\epsilon_0)^2} = \frac{4 \cdot 13.6 \mbox{ eV}}{n^2} = \frac{54.4...$
where $\mu \approx m_e$ is the reduced mass.
The energy of the photon is
$\begin{align*}\frac{hc}{\lambda} = \frac{4.14 \cdot 10^{-15}\mbox{ eV s} \cdot 3.00 \cdot 10^8 \mbox{ m/s}}{470 \cdot 10^{-9}...$
Thus,
$\begin{align*}\frac{54.4 \mbox{ eV}}{n^2} - \frac{54.4 \mbox{ eV}}{4^2} = 2.643 \mbox{ eV} \Rightarrow n = \sqrt{\frac{54.4 \...$
$\begin{align*}E_f = E_3 = \frac{-54.4 \mbox{ eV}}{n^2} = \boxed{-6.04} \mbox{ eV}\end{align*}$
Therefore, answer (A) is correct.

Solution to 1996 Problem 40