Q 12-12-061JEE MainJEE Main 2023 (31 Jan, Shift 1)Easy
For hydrogen atom, $\lambda_1$ and $\lambda_2$ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of $\lambda_1$ and $\lambda_2$ is $\dfrac{x}{32}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 27
$\dfrac1\lambda\propto\dfrac1{n_1^2}-\dfrac1{n_2^2}$.
Transition 1 ($3\to1$): $1-\dfrac19=\dfrac89$. Transition 2 ($2\to1$): $1-\dfrac14=\dfrac34$.
$$\frac{\lambda_1}{\lambda_2}=\frac{3/4}{8/9}=\frac{27}{32}$$
So $x=27$.
Solution by Sreeraj P, M.Sc Physics