The ratio of lowest energy in terms of wave numbers of Balmer and Lyman series of lines of atomic spectrum…
The ratio of lowest energy in terms of wave numbers of Balmer and Lyman series of lines of atomic spectrum of hydrogen is
5 : 27
27 : 5
20 : 27
27 : 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 $=5: 27$