Question

The ratio of the wave numbers for the highest energy transition of $e^{-}$ in Lyman and Balmer series of $H-$atom is:

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

Answer 1

Answer: (A)

4 : 1

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 $n_2=\infty$

$\therefore \frac{{\overline{v}}_1}{{\overline{v}}_2}=\frac{4}{1}$.