Question

The fast emission line on hydrogen atomic spectrum in the Balmer series appears at:

[$R=$ Rydberg constant]

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

Answer 1

Answer: (A)

$\frac{5R}{36}c{m}^{-1}$

Rydberg - Riz equation is :

$\overline{v}=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}](Where{n}_1=2,n_2=3)$

$\therefore \overline{v}=R_H\left[\frac{9-4}{64}\right]=\frac{5R}{36}c{m}^{-1}$