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JEE Main 2019 (12 Apr, Shift 2) · Transitions, Photon Emission & Excitation

Transitions, Photon Emission & Excitation question from JEE Main 2019 (12 Apr, Shift 2)

Q54JEE Main 2019 · 12 APR · SHIFT 2Medium

Transitions, Photon Emission & Excitation

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, $\frac{\lambda_{1}}{\lambda_{2}}$ , of the photons emitted in this process is:

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Answer: (d)

Third excited $n=4$ , second excited $n=3$ , first excited $n=2$ .
$\frac{1}{\lambda_1} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}$
$\frac{1}{\lambda_2} = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{20R}{144}$
$\frac{\lambda_1}{\lambda_2} = \frac{20}{7}$ , so the answer is (d).

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