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

Solution & Explanation

\textbf{Answer:} (B) $\frac{1}{4}$ \textbf{Solution:} Using $\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$. For $2\to 3$: $\frac{1}{{\lambda }_1}\propto \frac{1}{4}-\frac{1}{9}=\frac{5}{36}$. For $4\to 6$: $\frac{1}{{\lambda }_2}\propto \frac{1}{16}-\frac{1}{36}=\frac{5}{144}$. So $\frac{{\lambda }_1}{{\lambda }_2}=\frac{5/144}{5/36}=\frac{36}{144}=\frac{1}{4}$.