Question

The ratio of the wave numbers for the highest energy transition of electron in Lyman and Balmer series of H-atom is:

• A. 4 : 1
• B. 6: 1
• C. 9: 1
• D. 3: 1

Answer 1

The correct option is A 4 : 1

$\text{Lyman series}\frac{1}{{\lambda }_1}={\overline{v}}_1=R_H.1^2[\frac{1}{1^2}-\frac{1}{n_2^2}]$
Balmer series
$\frac{1}{{\lambda }_2}={\overline{v}}_2=R_H.1^2[\frac{1}{2^2}-\frac{1}{n_2^2}]$
For the highest energy transition:
For lyman series $n_2=\infty$
For Balmer series $n_2=\infty$
$\therefore \frac{{\overline{v}}_1}{{\overline{v}}_2}=\frac{4}{1}$