Question

A series of lines in the spectrum of atomic hydrogen lies at $656.46nm$, $486.27nm$, $439.17nm$ and $410.29nm$. What is the wavelength of the next line in this series? What is the ionisation energy of the atom when it is in the lower state of transition?

Answer 1

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

As we know, for Balmer series, $n₁=2$

$\lambda =410.29nm$

$1nm={10}^{-7}cm$

$\therefore \lambda =410.29\times {10}^{-7}$

$n_2=?$

$\frac{1}{2}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{1}{410.29\times {10}^{-7}}=109763[\frac{1}{2^2}-\frac{1}{n_2^2}]$

solving this, we get

$n_2=6$

The next line will be obtained during the jump of an electron from 7th to 2nd shell, so.

$\frac{1}{\lambda }=109673[\frac{1}{2^2}-\frac{1}{7^2}]$

$\frac{1}{\lambda }=109673[\frac{1}{4}-\frac{1}{49}]$

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

$\lambda =397.2nm$