What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He^+ spectrum?

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Question

Prashant Kotian · Accepted Answer

Correct answer: C. He^+ Balmer (n=4→2): 1/λ = R*4*(1/4 - 1/16) = 3R/4. For H: 1/λ = R*(1/1 - 1/4) = 3R/4. Transition is n=2 to n=1.

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