NEET (UG) 2025 — Question 43

Use the Rydberg formula for hydrogen: $\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. For the $n=2 \to n=3$ transition: $\frac{1}{\lambda_1} = R \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R \left(\frac{1}{4} - \frac{1}{9}\right) = R \left(\frac{9-4}{36}\right) = \frac{5R}{36}$. For the $n=4 \to n=6$ transition: $\frac{1}{\lambda_2} = R \left(\frac{1}{4^2} - \frac{1}{6^2}\right) = R \left(\frac{1}{16} - \frac{1}{36}\right) = R \left(\frac{9-4}{144}\right) = \frac{5R}{144}$. Take the ratio: $\frac{1/\lambda_1}{1/\lambda_2} = \frac{5R/36}{5R/144} = \frac{144}{36} = 4$. Thus $\frac{\lambda_2}{\lambda_1} = 4$, so $\frac{\lambda_1}{\lambda_2} = \frac{1}{4}$. The required ratio is $\frac{1}{4}$, which corresponds to option D.