The ratio of the wavelengths of the light absorbed by a hydrogen atom when it undergoes $n=2 \rightarrow n=3$ and $n=4 \rightarrow n=6$ transitions, respectively, is:

Solution & Explanation

\textbf{Answer:} (B) $\dfrac{1}{4}$ \textbf{Solution:} Using $\dfrac{1}{\lambda} = R Z^{2}\left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right)$. For $2\rightarrow3$: $\dfrac{1}{\lambda_1} \propto \dfrac{1}{4} - \dfrac{1}{9} = \dfrac{5}{36}$. For $4\rightarrow6$: $\dfrac{1}{\lambda_2} \propto \dfrac{1}{16} - \dfrac{1}{36} = \dfrac{5}{144}$. So $\dfrac{\lambda_1}{\lambda_2} = \dfrac{5/144}{5/36} = \dfrac{36}{144} = \dfrac{1}{4}$.