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The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561 Å .$ The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

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$1215 Å$

$1640 Å$

$2430 Å$

$4687 Å$

answer is A.

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Detailed Solution

$\frac{1}{6561} = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}$

$\frac{1}{λ} = 4 R (\frac{1}{4} - \frac{1}{16}) = \frac{3 R \times 4}{16}$

$λ = 1215 Å$

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