Solution to 1996 Problem 40

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}...
\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.

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