Q41 / 180 · NEET 2025 · Chemistry
The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes $\mathrm{n}=2 \rightarrow \mathrm{n}=3$ and $\mathrm{n}=4 \rightarrow \mathrm{n}=6$ transitions, respectively, is
- A$\frac{1}{36}$
- B$\frac{1}{16}$
- C$\frac{1}{9}$
- Correct answerD$\frac{1}{4}$
Solution
$\frac{1}{\lambda}=\mathrm{RZ}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
For $2 \rightarrow 3 \Rightarrow \frac{1}{\lambda_1}=\mathrm{R} \times(1)^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \ldots(1)$
For $4 \rightarrow 6 \Rightarrow \frac{1}{\lambda_2}=\mathrm{R} \times(1)^2\left(\frac{1}{4^2}-\frac{1}{6^2}\right) \ldots(2)$
Divide (1) & (2)
$\begin{aligned}
& \frac{\lambda_2}{\lambda_1}=\frac{\frac{1}{2^2}-\frac{1}{3^2}}{\frac{1}{4^2}-\frac{1}{6^2}} \\
& \frac{\lambda_2}{\lambda_1}=\frac{4}{1} \\
& \therefore \frac{\lambda_1}{\lambda_2}=\frac{1}{4}
\end{aligned}$
Hence, the correct answer is option: 'D'.