A parallel beam of light of wavelength 100 nm passes through a...

The energy of the incident photon is given by:

$E = \dfrac{hc}{\lambda} = \dfrac{1242 \text{ eV nm}}{100 \text{ nm}} = 12.42 \text{ eV}$

The energy required to excite a hydrogen atom from the ground state ($n=1$) to higher energy states is:

$\Delta E_{1 \rightarrow 2} = 13.6 \left(1 - \dfrac{1}{2^2}\right) = 10.2 \text{ eV}$

$\Delta E_{1 \rightarrow 3} = 13.6 \left(1 - \dfrac{1}{3^2}\right) = 12.09 \text{ eV}$

$\Delta E_{1 \rightarrow 4} = 13.6 \left(1 - \dfrac{1}{4^2}\right) = 12.75 \text{ eV}$

Since the incident photon energy ($12.42 \text{ eV}$) is greater than $12.09 \text{ eV}$ but less than $12.75 \text{ eV}$, the hydrogen atoms can be excited to the $n=2$ and $n=3$ states. The remaining energy is carried by photons moving in the forward direction, which are not detected by the perpendicular detector.

The excited atoms will de-excite and emit photons in all directions. The possible emission transitions are $3 \rightarrow 1$, $2 \rightarrow 1$, and $3 \rightarrow 2$.

The wavelengths of the emitted photons are calculated as follows:

For $3 \rightarrow 1$: $\lambda_{31} = \dfrac{1242}{12.09} \approx 103 \text{ nm}$

For $2 \rightarrow 1$: $\lambda_{21} = \dfrac{1242}{10.2} \approx 121 \text{ nm}$

For $3 \rightarrow 2$: $\lambda_{32} = \dfrac{1242}{1.89} \approx 657 \text{ nm}$ (closest to $654 \text{ nm}$ in the given options)

Therefore, the detector will register wavelengths of $103 \text{ nm}$, $121 \text{ nm}$, and $654 \text{ nm}$.

Answer: $103 \text{ nm}$, $121 \text{ nm}$, $654 \text{ nm}$