Q 12-12-119JEE MainJEE Main 2019 (12 Apr, Shift 2)Easy
The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths $\dfrac{\lambda_1}{\lambda_2}$ of the photons emitted in this process is:
Answer: (B) $\dfrac{20}{7}$
Transitions $4 \to 3$ and $3 \to 2$:
$$\frac1{\lambda_1} \propto \frac19 - \frac1{16} = \frac{7}{144},\qquad \frac1{\lambda_2} \propto \frac14 - \frac19 = \frac{20}{144}$$
$$\frac{\lambda_1}{\lambda_2} = \frac{20}{7}$$
Solution by Sreeraj P, M.Sc Physics