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In a hydrogen atom, an electron transitions from the 4th excited state to the 2nd excited state. What is the wavelength of the emitted photon in terms of the Rydberg constant R?

1. 225/16R
2. 225/4R
3. 100/21R
4. 100/4R

Correct answer: 225/16R

Solution

The 4th excited state corresponds to n=5 and the 2nd excited state to n=3. The Rydberg formula gives 1/lambda = R*(1/3² - 1/5²) = R*(1/9 - 1/25) = R*(16/225). Therefore lambda = 225/(16R).

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