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A hydrogen atom in an excited state emits a photon of wavelength lambda and transitions to the ground state (n = 1). If R is the Rydberg constant, the principal quantum number n of the excited state is:

1. sqrt(lambda * R * (lambda * R - 1))
2. sqrt(lambda * R / (lambda * R - 1))
3. sqrt((lambda * R - 1) * lambda * R)
4. sqrt((lambda * R - 1) / (lambda * R))

Correct answer: sqrt(lambda * R / (lambda * R - 1))

Solution

Applying the Rydberg formula for emission to ground state: 1/lambda = R(1 - 1/n²). Rearranging yields n² = lambda*R/(lambda*R - 1), so n = sqrt(lambda*R/(lambda*R - 1)).

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