Question

The first emission line of Balmer series in H spectrum has wave number equal to ?

• A. $\frac{9R_H}{400}c{m}^{-1}$
• B. $\frac{7R_H}{144}c{m}^{-1}$
• C. $\frac{3R_H}{4}c{m}^{-1}$
• D. $\frac{5R_H}{36}c{m}^{-1}$

Answer 1

The correct option is D $\frac{5R_H}{36}c{m}^{-1}$

Given : The first emission line of Balmer series in H spectrum has wave number equal to ?

Solution :

$\overrightarrow{v}=\frac{1}{\lambda }$ = $R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

For Balmer series $n_1=2$ and $n_2=3$ as for first emmision.

$\overrightarrow{v}=R_H[\frac{1}{4}-\frac{1}{9}]$ = $\frac{5R_H}{36}c{m}^{-1}$

The Correct Opt = D