Physics - Modern Physics Question with Solution | TestHub

The wavelength of different members of Balmer series are given by

$\frac{1}{\lambda }$$=R_{\mathrm{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.${\mathrm{H}}_{\alpha }$line (or longest wavelength )

$\frac{1}{{\lambda }_1}$$=R_{\mathrm{H}}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$

$=1.097\times {10}^7\left(\frac{5}{36}\right)$

or${\lambda }_1=\frac{36}{5\times \text{1.097}\times 10^7}=\text{6.563}\times 10^{-7}\mathrm{ m}$

$n=6563\mathrm{ \AA }$

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${\lambda }_{\infty }=\frac{4}{\text{1.097}\times 10^7}=\text{3.646}\times 10^{-7}\text{ m}$

$=3646\mathrm{ \AA }$

Only$4861\mathrm{ \AA }$is between the first and last line of the Balmer series.