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Question: Answered & Verified by Expert
The ratio of lowest energy in terms of wave numbers of Balmer and Lyman series of lines of atomic spectrum of hydrogen is
ChemistryStructure of AtomAP EAMCETAP EAMCET 2018 (22 Apr Shift 2)
Options:
  • A 5 : 27
  • B 27 : 5
  • C 20 : 27
  • D 27 : 2
Solution:
2818 Upvotes Verified Answer
The correct answer is: 5 : 27
(a) $\because$ Wave number $(v)=R \cdot Z^2\left\lceil\frac{1}{n_{\mathrm{L}^2}}-\frac{1}{n_{\mathrm{H}^2}}\right\rfloor$
Wave number for lowest energy for Balmer series $\left(n_{\mathrm{L}}=2, n_{\mathrm{H}}=3\right)$
$$
v=R\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5}{36} R
$$
Wave number for lowest energy for Lyman series:
$$
\left(n_{\mathrm{L}}=1, n_{\mathrm{H}}=2\right) \Rightarrow v=R\left[1-\frac{1}{4}\right]=\frac{3}{4} R
$$
Thus, ratio of Balmer/Lyman is
$$
=\frac{R .5 / 36}{R .3 / 4}=\frac{5 \times 4}{3 \times 36}=\frac{5}{27}
$$
Hence ratio $=5: 27$

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