Question

The first emission line in the H-atom spectrum in the Balmer series will have a wave number of:

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

Answer 1

Answer: (A)

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

$\frac{1}{\lambda }=\overline{v}=R{z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2});$

For first emission lione of Balmer series $n_1=2,n_2=3$

Then, by using Rydberg's formula

$\overline{v}=R(\frac{1}{4}-\frac{1}{9})=\frac{5R}{36}c{m}^{-1}$

Hence, the correct option is $A$