A collection of hydrogen atoms is prepared in the n = 4 state....

The possible transitions for a hydrogen atom returning from the $n=4$ state to the $n=2$ state are:
$n=4 \rightarrow n=2$
$n=4 \rightarrow n=3$
$n=3 \rightarrow n=2$

Using the Rydberg formula for the wavelength of emitted radiation:
\dfrac{1}{\lambda} = R \left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)
where $R \approx 1.097 \times 10^7 \text{ m}^{-1}$.

For the $n=4 \rightarrow n=2$ transition:
\dfrac{1}{\lambda_1} =R \left(\dfrac{1}{2^2} - \dfrac{1}{4^2} \right) =R \left(\dfrac{1}{4} - \dfrac{1}{16} \right) = \dfrac{3R}{16}
$\lambda_1 =$\dfrac{16}{3R} \approx$486 \text{ nm}$

For the $n=3 \rightarrow n=2$ transition:
\dfrac{1}{\lambda_2} =R \left(\dfrac{1}{2^2} - \dfrac{1}{3^2} \right) =R \left(\dfrac{1}{4} - \dfrac{1}{9} \right) = \dfrac{5R}{36}
$\lambda_2 =$\dfrac{36}{5R} \approx$656 \text{ nm}$

For the $n=4 \rightarrow n=3$ transition:
\dfrac{1}{\lambda_3} =R \left(\dfrac{1}{3^2} - \dfrac{1}{4^2} \right) =R \left(\dfrac{1}{9} - \dfrac{1}{16} \right) = \dfrac{7R}{144}
$\lambda_3 =$\dfrac{144}{7R} \approx$1875 \text{ nm}$

Comparing the calculated wavelengths ($486 \text{ nm}$, $656 \text{ nm}$, $1875 \text{ nm}$) with the given options, the closest values are $487 \text{ nm}$, $654 \text{ nm}$, and $1910 \text{ nm}$.