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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                     &#1

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