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For the He⁺ ion, the light emitted in the transition from n=4 to n=2 has the same frequency as which transition in the hydrogen atom?

1. n=2 to n=1
2. n=3 to n=2
3. n=4 to n=3
4. n=3 to n=1

Correct answer: n=2 to n=1

Solution

Frequency is proportional to Z^2(1/n_f^2 - 1/n_i^2). For He+ (Z=2), 4->2 gives 4*(1/4 - 1/16) = 3/4. For hydrogen, 2->1 gives (1 - 1/4) = 3/4, which is the same. So it matches the n=2 to n=1 transition.

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