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

Answer 1

The Correct Option is 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 $n _{2}=3$.