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The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is:

Options

  1. A4: 1
  2. B1: 2
  3. C1: 4
  4. D2: 1

Correct answer

A. 4: 1

Step-by-step solution

Wavelength of H -atom is 1 = Rz ^2 ( 1 n ₁^2 - 1 n ₂^2 ) Shortest wavelength for Balmer series: 1 _B = Rz ^2 ( 1 2^2 - 1 ) Shortest wavelength for Lyman series: 1 _ L = Rz ^2 ( 1 1^2 - 1 ) ....(ii) Dividing eq. (i) and (ii), _ A : ₁=4: 1

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