Question:
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 :
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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Third excited state is \(n=4\), second excited \(n=3\), first excited \(n=2\). Use \(\frac{1}{\lambda} = R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\) for each jump.
Updated On: Nov 11, 2026
- \(\frac{22}{5}\)
- \(\frac{7}{5}\)
- \(\frac{9}{7}\)
- \(\frac{20}{7}\)
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The Correct Option is D
Solution and Explanation
The transitions are \(n=4\to3\) (wavelength \(\lambda_1\)) and \(n=3\to2\) (wavelength \(\lambda_2\)).
\(\frac{1}{\lambda_1} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144}\).
\(\frac{1}{\lambda_2} = R\left(\frac{1}{4}-\frac{1}{9}\right) = \frac{5R}{36}\).
\(\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{5/36}{7/144} = \frac{5\times144}{36\times7} = \frac{20}{7}\).
Answer: (4) \(\frac{20}{7}\)
\(\frac{1}{\lambda_1} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144}\).
\(\frac{1}{\lambda_2} = R\left(\frac{1}{4}-\frac{1}{9}\right) = \frac{5R}{36}\).
\(\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{5/36}{7/144} = \frac{5\times144}{36\times7} = \frac{20}{7}\).
Answer: (4) \(\frac{20}{7}\)
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