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An electron in a hydrogen atom first transitions from the third excited state (n=4) to the second excited state (n=3), and then from the second excited state (n=3) to the first excited state (n=2). Find the ratio of the wavelengths lambda1/lambda2 of the photons emitted in these two transitions.

1. 9/7
2. 7/5
3. 27/5
4. 20/7

Correct answer: 20/7

Solution

Third excited state = n=4; second excited state = n=3; first excited state = n=2. Transition 1 (n=4 to n=3): 1/lambda1 = R*(1/3² - 1/4²) = R*(1/9 - 1/16) = R*(16-9)/144 = 7R/144. So lambda1 = 144/(7R). Transition 2 (n=3 to n=2): 1/lambda2 = R*(1/2² - 1/3²) = R*(1/4 - 1/9) = R*(9-4)/36 = 5R/36. So lambda2 = 36/(5R). Ratio: lambda1/lambda2 = [144/(7R)] / [36/(5R)] = (144*5)/(7*36) = 720/252 = 20/7.

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