Question

Which of the following lines of the $H$ -atom spectrum belongs to the Balmer series?

• A. $1025Å$
• B. $1218Å$
• C. $4861Å$
• D. $18751Å$

Answer 1

Answer: (C)

The wavelength of different members of Balmer series are given by
$\frac{1}{\lambda }$ $=R_H\left[\frac{1}{2^2}-\frac{1}{n_i^2}\right]$ , where $n_i=\text{3, 4, 5....}$
The first member of Balmer series (H_α) corresponds to $n_i=\text{3}$
It has maximum energy and hence the longest wavelength. $H_{\alpha }$ line (or longest wavelength )
$\frac{1}{{\lambda }_1}$ $=R_H\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$
$=1.097\times 10^7\left(\frac{5}{36}\right)$
or ${\lambda }_1=\frac{36}{5\times \text{1.097}\times 10^7}=\text{6.563}\times 10^{-7}m$
$n=6563\mathring{A}$
The wavelength of the Balmer series limit corresponds to $n_i=\infty$ and has got shortest wavelength.
Therefore, wavelength of Balmer series limit is given by
$\frac{1}{{\lambda }_{\infty }}={\text{R}}_{\text{H}}\left[\frac{1}{2^2}-\frac{1}{{\infty }^2}\right]=\text{1.097}\times 10^7\times \frac{1}{4}$
or ${\lambda }_{\infty }=\frac{4}{\text{1.097}\times 10^7}=\text{3.646}\times 10^{-7}\text{ m}$
$=3646\mathring{A}$
Only $4861\mathring{A}$ is between the first and last line of the Balmer series.