Question

For Balmer series in the spectrum of atomic hydrogen, the wave number of each line is given by $\bar{\upsilon }=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$ where $R_H$ is a constant and $n_1$and $n_2$ are integers. Which of the following statement(s) is (are) correct?
1. As wavelength decreases, the lines in the series converge
2. The integer $n_1$ is equal to $2$
3. The ionisation energy of hydrogen can be calculated from the wave number of these lines
4. The line of longest wavelength corresponds to $n_2=3$

• A. $1,2$ and $3$
• B. $2,3$ and $4$
• C. $1,2$ and $4$
• D. $2$ and $4$ only
• E. $2$ only

Answer 1

Answer: (C)

$\overline{v}=\frac{1}{\lambda }=R_H\left[\frac{1}{2^2}-\frac{1}{n_2^2}\right]$
Here $n=2$ since Balmer series, the transition is between some energy level $n_2$ and $n_1=2.$
As the wave length decreases, the wave number increases.
This implies that the difference $\frac{1}{2^2}-\frac{1}{n_2^2}$ also increases or ultimately $m_2$ increases.
As $n_2$ increases, the lines in the series coverge.
Ionization energy of hydrogen can be calculated from wave number of lines in lyman series as the electron is present in $n=1$.
Longest wave length corresponds to minimum difference between
$\frac{1}{2^2}$ and $\frac{1}{n_2^2}$. This is for $n_2=3$.