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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 λ1/λ2 of the photons emitted in this process is:

1. 22/5
2. 7/5
3. 9/7
4. 20/7

Correct answer: 20/7

Solution

3rd excited (n=4) -> 2nd excited (n=3): 1/lambda1 = R(1/9 - 1/16) = 7R/144. 2nd excited (n=3) -> 1st excited (n=2): 1/lambda2 = R(1/4 - 1/9) = 5R/36. lambda1/lambda2 = (5/36)/(7/144) = 20/7.

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