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

Energy of the incident photon is given by:

$E = \dfrac{hc}{\lambda} = \dfrac{1240 \text{ eV nm}}{100 \text{ nm}} = 12.4 \text{ eV}$

The energy levels of a hydrogen atom are $E_n = -\dfrac{13.6}{n^2} \text{ eV}$.

The energy required to excite a hydrogen atom from the ground state ($n=1$) to higher states is:
For $n=2$: $\Delta E_{12} = 13.6 \left(1 - \dfrac{1}{4}\right) = 10.2 \text{ eV}$
For $n=3$: $\Delta E_{13} = 13.6 \left(1 - \dfrac{1}{9}\right) \approx 12.08 \text{ eV}$
For $n=4$: $\Delta E_{14} = 13.6 \left(1 - \dfrac{1}{16}\right) = 12.75 \text{ eV}$

Since the incident photon energy ($12.4 \text{ eV}$) is less than $12.75 \text{ eV}$, it can only excite the atom to $n=2$ or $n=3$.

When the atom is excited to $n=2$, the remaining energy of the photon is:
$E_2' = 12.4 - 10.2 = 2.2 \text{ eV}$
The wavelength of this transmitted photon is:
$\lambda_2' = \dfrac{1240}{2.2} \approx 563 \text{ nm} \approx 560 \text{ nm}$

When the atom is excited to $n=3$, the remaining energy of the photon is:
$E_3' = 12.4 - 12.08 = 0.32 \text{ eV}$
The wavelength of this transmitted photon is:
$\lambda_3' = \dfrac{1240}{0.32} = 3875 \text{ nm} \approx 3880 \text{ nm}$

The unabsorbed incident photons will also pass through the gas without interacting, so the original $100 \text{ nm}$ wavelength will still be present in the transmitted beam.

Therefore, the wavelengths observed in the transmitted beam are $100 \text{ nm}$, $560 \text{ nm}$, and $3880 \text{ nm}$.

Answer: $100 \text{ nm}$, $560 \text{ nm}$, $3880 \text{ nm}$