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AP EAMCET201822 Apr 2018Evening ShiftChemistryStructure of AtomActual

The ratio of lowest energy in terms of wave numbers of Balmer and Lyman series of lines of atomic spectrum of hydrogen is

Options

  1. A5 : 27
  2. B27 : 5
  3. C20 : 27
  4. D27 : 2

Correct answer

A. 5 : 27

Step-by-step solution

(a) Wave number (v)=R Z^2 1 n_ L ^2 - 1 n_ H ^2 Wave number for lowest energy for Balmer series (n_ L =2, n_ H =3 ) v=R [ 1 4 - 1 9 ]= 5 36 R Wave number for lowest energy for Lyman series: (n_ L =1, n_ H =2 ) v=R [1- 1 4 ]= 3 4 R Thus, ratio of Balmer/Lyman is = R .5 / 36 R .3 / 4 = 5 4 3 36 = 5 27 Hence ratio =5: 27

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