Q 12-12-065JEE MainJEE Main 2023 (25 Jan, Shift 1)Medium
The wavelength of the radiation emitted is $\lambda_0$ when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be $\dfrac{20}{x}\lambda_0$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 27
$3\to2$: $\dfrac1{\lambda_0}\propto\dfrac14-\dfrac19=\dfrac5{36}$. $\ 4\to2$: $\dfrac1\lambda\propto\dfrac14-\dfrac1{16}=\dfrac3{16}$.
$$\lambda=\lambda_0\times\frac{5/36}{3/16}=\frac{20}{27}\lambda_0$$
So $x=27$.
Solution by Sreeraj P, M.Sc Physics