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Q.
Which of the following lines of the $H$ -atom spectrum belongs to the Balmer series?
NTA AbhyasNTA Abhyas 2022
Solution:
The wavelength of different members of Balmer series is 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 ( $H_{\alpha }$ ) corresponds to $n _{i } = \text{3}$
It has maximum energy and hence the longest wavelength. $H_{\alpha }$ line (or longest wavelength )
$\frac{1}{\lambda _{1}}$ $=R_{H}\left[\frac{1}{2^{2}} - \frac{1}{3^{2}}\right]$
$=1.097\times \, \left(10\right)^{7}\left(\frac{5}{36}\right)$
or $\lambda _{1} = \frac{3 6}{5 \times \text{1.097} \times 1 0^{7}} = \text{6.563} \times 1 0^{- 7} m$
$n = 6 5 6 3 \mathring{A} $
The wavelength of the Balmer series limit corresponds to $n _{i } = \infty $ and has got the shortest wavelength.
Therefore, the wavelength of Balmer series limit is given by
$\frac{1}{\lambda _{\infty }} = \text{R}_{\text{H}} \left[\frac{1}{2^{2}} - \frac{1}{\infty ^{2}}\right] = \text{1.097} \times 1 0^{7} \times \frac{1}{4}$
or $\lambda _{\infty } = \frac{4}{\text{1.097} \times 1 0^{7}} = \text{3.646} \times 1 0^{- 7} \text{ m}$
$= 3 6 4 6 \mathring{A} $
Only $4861 \mathring{A} $ is between the first and last line of the Balmer series.