Q2 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Chemistry
The ratio of lowest energy in terms of wave numbers of Balmer and Lyman series of lines of atomic spectrum of hydrogen is
- Correct answerA5 : 27
- B27 : 5
- C20 : 27
- D27 : 2
Solution
(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 $=5: 27$
Hence, the correct answer is option: 'A'.