Question

Find out the wavelength of the next line in the series having lines of spectrum of $H$-atom of wavelengths $6565\mathring{A}$, $4863\mathring{A}$, $4342\mathring{A}$ and $4103\mathring{A}$.

Answer 1

The given series lies in the visible region and thus appears to be Balmer series.

Given:

$n_1$ = 2

$\lambda =410.3\times {10}^{-7}$ cm

$n_2$=?

$\frac{1}{\lambda }=R_H\times Z^2$ [$\frac{1}{n_1^2}$ $-\frac{1}{n_2^2}$]

$1410.3\times {10}^{-7}=109678[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\therefore n_2=6$

Next line will be obtained during the jump of electron from 7th to 2nd shell-

$\frac{1}{\lambda }=R_H[\frac{1}{2^2}-\frac{1}{7^2}]$

$=109678\times 1^2[\frac{1}{4}-\frac{1}{49}]$

$\lambda =397.2\times {10}^{-7}$ cm

$\lambda =3972\mathring{A}$

Hence, the next wavelength in this series will be 3972 $\mathring{A}$.