Q 12-12-068JEE MainJEE Main 2023 (15 Apr, Shift 1)Medium
As per given figure $A$, $B$ and $C$ are the first, second and third excited energy levels of hydrogen atom respectively. If the ratio of the two wavelengths $\left(\text{i.e. }\dfrac{\lambda_1}{\lambda_2}\right)$ is $\dfrac7{4n}$, then the value of $n$ will be ______.
Numerical value type. Enter your answer.
Answer: 5
$A$, $B$, $C$ are $n=2,3,4$.
$\lambda_1$ ($3\to2$): $\dfrac1{\lambda_1}\propto\dfrac14-\dfrac19=\dfrac5{36}$. $\ \lambda_2$ ($4\to3$): $\dfrac1{\lambda_2}\propto\dfrac19-\dfrac1{16}=\dfrac7{144}$.
$$\frac{\lambda_1}{\lambda_2}=\frac{7/144}{5/36}=\frac{7}{20}=\frac7{4\times5}$$
So $n=5$.
Solution by Sreeraj P, M.Sc Physics