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The wavelength of the limiting line of the Lyman series for the He⁺ ion is x angstroms. What is the wavelength of the limiting line of the Balmer series for the Li²+ ion?

1. 9x/4 Å
2. 16x/9 Å
3. 5x/4 Å
4. 4x/7 Å

Correct answer: 16x/9 Å

Solution

Limiting line means n2 -> infinity. Wavelength formula: 1/lambda = R*Z²/n1². For He⁺ (Z=2), Lyman series (n1=1): 1/x = R*(4)/1 = 4R => R = 1/(4x). For Li²+ (Z=3), Balmer series (n1=2): 1/lambda = R*(9)/4 = (1/(4x))*(9/4) = 9/(16x). So lambda = 16x/9 Å.

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