A $He^+$ ion undergoes a transition to the ground state, emitting two photons in succession with wav

Question

A $He^+$ ion undergoes a transition to the ground state, emitting two photons in succession with wavelengths 108.5 nm and 30.4 nm. Taking the Rydberg constant $R$ and Planck's constant such that $hc \approx 1240 	ext{ eV nm}$, what is the principal quantum number $n$ of the initial excited state?

Professor PC Thomas Classes · Answer

Energy of the second photon: $E_2 = 1240 / 30.4 \approx 40.8 	ext{ eV}$.
The ground state energy of $He^+$ ($Z=2$) is $E_1 = -13.6 	imes 2^2 = -54.4 	ext{ eV}$.
The energy of the $n=2$ state is $E_2 = -54.4 / 4 = -13.6 	ext{ eV}$.
Notice that $-13.6 - (-54.4) = 40.8 	ext{ eV}$. Thus, the second transition is from $n=2$ to $n=1$.
Energy of the first photon: $E_1 = 1240 / 108.5 \approx 11.4 	ext{ eV}$.
Let the initial state be $n$. The transition is from $n$ to $n=2$.
$\Delta E = -54.4/n^2 - (-13.6) = 11.4 	ext{ eV}$
$13.6 - 11.4 = 54.4 / n^2 \implies 2.2 = 54.4 / n^2 \implies n^2 \approx 24.7$
Thus, $n = 5$.