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 $=5: 27$