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, $\frac{λ_1}{λ_2}$, of the photons emitted in this process is:
Answer & explanation
Correct answer: option 1
Third excited state(n=4) to second excited state (n=3) $\Rightarrow \frac{1}{\lambda_1} = R(\frac{1}{3^2} - \frac{1}{4^2}) = \frac{7R}{144}$
Second excited state(n=3) to first excited state (n=2)
$\Rightarrow \frac{1}{\lambda_2} = R(\frac{1}{2^2} - \frac{1}{3^2}) = \frac{5R}{36}$
$\Rightarrow \frac{\lambda_1}{\lambda_2} = \frac{5R/36}{7R/144} = \frac{20}{7}$