A light beam with wavelengths distributed uniformly between 450 nm...

The least intensity in the transmitted beam corresponds to the wavelength that is absorbed by the hydrogen gas.

The given wavelength range is $450 \text{ nm}$ to $550 \text{ nm}$, which falls in the visible region. Absorption in the visible region corresponds to the Balmer series, where the electron transitions from the $n = 2$ state to a higher energy level.

The energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by:

$E_n = -\dfrac{13.6}{n^2} \text{ eV}$

For the $n = 2$ state, $E_2 = -3.4 \text{ eV}$.

For the $n = 3$ state, $E_3 = -1.51 \text{ eV}$. The energy difference is $\Delta E = 3.4 - 1.51 = 1.89 \text{ eV}$, which corresponds to $\lambda = \dfrac{1242}{1.89} \approx 657 \text{ nm}$ (outside the given range).

For the $n = 4$ state, $E_4 = -0.85 \text{ eV}$. The energy difference for the transition from $n = 2$ to $n = 4$ is:

$\Delta E = E_4 - E_2 = -0.85 - (-3.4) = 2.55 \text{ eV}$

The corresponding wavelength is given by:

$\lambda = \dfrac{1242 \text{ eV nm}}{\Delta E} = \dfrac{1242}{2.55} \approx 487 \text{ nm}$

Since $487 \text{ nm}$ lies within the given range of $450 \text{ nm}$ to $550 \text{ nm}$, this wavelength will be absorbed by the hydrogen gas, resulting in the least intensity in the transmitted beam.