Question

To what series does the spectral lines of atomic hydrogen belong of its wave number is equal to the difference between the wave numbers of the following two lines of the Balmer series $486.1nm$ and $410.2nm$? What is the wavelength of this line?

• A. Brackett, $\lambda =125\times {10}^{-4}cm$
• B. Paschen, $\lambda =26.3\times {10}^{-4}cm$
• C. Paschen, $\lambda =2.63\times {10}^{-4}cm$
• D. Lyman, $\lambda =9.1\times {10}^{-5}cm$

Answer 1

The correct option is C Paschen, $\lambda =2.63\times {10}^{-4}cm$

Given that; ${\lambda }_1=486.1\times {10}^{-9}m=486.1\times {10}^{-7}cm$
${\lambda }_2=410.2\times {10}^{-9}m=410.2\times {10}^{-7}cm$
and $\overline{v}={\overline{v}}_2-{\overline{v}}_1=[\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}]$
$=R_H[\frac{1}{2^2}-\frac{1}{n_2^2}]-R_H[\frac{1}{2^2}-\frac{1}{n_1^2}]$
$\overline{v}=R_H[\frac{1}{n^2}-\frac{1}{n_2^2}]$
For $1$ line of Balmer series:
$\frac{1}{{\lambda }_1}=R_H[\frac{1}{2^2}-\frac{1}{n_1^2}]=109678[\frac{1}{2^2}-\frac{1}{n_1^2}]$
or $\frac{1}{486.1\times {10}^{-7}}=109678[\frac{1}{2^2}-\frac{1}{n_1^2}]$
$\therefore n_1=4$
For $II$ line of Balmer series:
$\frac{1}{{\lambda }_1}=R_H[\frac{1}{2^2}-\frac{1}{n_2^2}]=109678[\frac{1}{2^2}-\frac{1}{n_2^2}]$
or $\frac{1}{410.2\times {10}^{-7}}=109678[\frac{1}{2^2}-\frac{1}{n_2^2}]$
$\therefore n_2=6$
Thus, given electronic transition occurs from $6th$ to $4th$ shell,i.e.,
Paschen series. Also by Eq.(1),
$\overline{v}=\frac{1}{\lambda }=109678[\frac{1}{4^2}-\frac{1}{6^2}]$
$\therefore \lambda =2.63\times {10}^{-4}cm$