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The shortest wavelength of H atoms in the Lyman series is λ₁. The longest wavelength in the Balmer series of He⁺ is:

1. 9λ₁/5
2. 27λ₁/5
3. 36λ₁/5
4. 5λ₁/9

Correct answer: 9λ₁/5

Solution

Shortest Lyman wavelength of H: 1/lambda1 = R(1) so lambda1 = 1/R. Longest Balmer line of He+ (Z=2) is 3->2: 1/lambda = R*4*(1/4 - 1/9) = 5R/9, so lambda = 9/(5R) = (9/5)lambda1.

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