A& sample of atomic hydrogen gas is irradiated by a beam of...

The energy of a& photon corresponding to $\lambda = 100nm$ is,

$\dfrac{1242\text{eV nm}}{100\text{nm}} = 12 \cdot 42\text{eV}$

Similarly, the corresponding to $\lambda = 200nm$ is $6 . 21eV$.

The energy needed to take the atom from the ground state to the first excited state is

${E}_{2} - {E}_{1} = 13 . 6eV - 3 . 4eV = 10 . 2eV$

The energy needed to take the atom from the ground state to the second excited state is

${E}_{3} - {E}_{1} = 13 . 6eV - 1 . 5eV = 12 . 1eV$

The energy needed to take the atom from the ground state to the third excited state is

${E}_{4} - {E}_{1} = 13 . 6eV - 0 . 85eV = 12 . 75eV$, and so on.

Thus, $10 . 2eV$ photons and $12 . 1eV$ photons have a large probability of being absorbed from the given range of $6 . 21eV$ to $12 . 42eV$.

The corresponding wavelengths are

${λ}_{1} = \dfrac{1 2 4 2 \text{eV} \text{nm}}{1 0 \cdot 2 \text{eV}} = 1 2 2 \text{nm}$

${λ}_{2} = \dfrac{1 2 4 2 \text{eV} \text{nm}}{1 2 \cdot 1 \text{eV}} = 1 0 3 \text{nm}$
These wavelengths will have low intensity in the transmitted beam.