Question

The frequency of one of the lines in Paschen series of hydrogen atom is $2.340\times {10}^{11}$ Hz. The quantum number $n_2$ which produces this transition is:

• A. 6
• B. 5
• C. 4
• D. 3

Answer 1

The correct option is D

3

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

=$\frac{1}{\lambda }$ = $R_H[\frac{1}{3^2}-\frac{1}{n_2^2}]$ = $n_2$ = $3$

For paschen series.