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If the series limit wavelength of the Lyman series of hydrogen atom is 912 Å , then the series limit wavelength of the Balmer series of the hydrogen atom is

Options

  1. A912 Å
  2. B1824 Å
  3. C3648 Å
  4. D456 Å

Correct answer

C. 3648 Å

Step-by-step solution

For series limit of Balmer series, n 2 = 2 ,   n 1 = ∞ 1 λ = R 1 n 2 2 - 1 n 1 2 = 1 2 2 - 1 ∞ 2 = R 4 ∴ λ = 4 R = 4 10967800   m = 4 × 912 × 10 - 10   m = 3648   Å

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