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A spectral line in the emission spectrum of Li2+ has the same wavelength as the second line of the Balmer series of hydrogen. If this Li2+ line arises from the transition n = 12 to n = x, what is the value of x?

1. 8
2. 6
3. 7
4. 5

Correct answer: 6

Solution

Second Balmer line of H is 4->2: 1/lambda = R(1/4 - 1/16) = 3R/16. For Li2+ (Z=3): 1/lambda = 9R(1/x^2 - 1/12^2). Setting 9(1/x^2 - 1/144) = 3/16 gives 1/x^2 = 1/36, so x = 6.

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