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The wavelength of the first spectral line in the Balmer series of the hydrogen atom is 6561 A ∘ . The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

Options

  1. A1215 A ∘
  2. B1640 A ∘
  3. C2430 A ∘
  4. D4687 A ∘

Correct answer

A. 1215 A ∘

Step-by-step solution

For hydrogen or hydrogen type atoms 1 λ = R Z 2 1 n f 2 - 1 n i 2 In the transition from n i → n f ∴ λ ∝ 1 Z 2 1 n f 2 - 1 n i 2 ∴ λ 2 λ 1 = Z 1 2 1 n f 2 - 1 n i 2 1 Z 2 2 1 n f 2 - 1 n i 2 2 λ 2 = λ 1 Z 1 2 1 n f 2 - 1 n i 2 1 Z 2 2 1 n f 2 - 1 n i 2 2 Substituting the values, we have = ( 6561 ) 1 2 1 2 2 - 1 3 2 2 2 1 2 2 - 1 4 2 = 1215 A ∘

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