Question

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

• A. 5 : 27
• B. 27 : 5
• C. 20 : 27
• D. 27 : 2

Answer 1

Answer: (A)

$\because$ Wave number $(v)=R\cdot Z^2\left[\frac{1}{n_{L^2}}-\frac{1}{n_{H^2}}\right]$

Wave number for lowest energy for Balmer series $\left(n_L=2,n_H=3\right)$

$v=R\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5}{36}R$

Wave number for lowest energy for Lyman series:

$\left(n_L=1,n_H=2\right)$

$\Rightarrow v=R\left[1-\frac{1}{4}\right]=\frac{3}{4}R$

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 $=5:27$