Question

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes a transition from n = 4 to n = 2?

Answer 1

Given.

transition From $n=4$ to $n=2$

To find

wave length $\lambda =?$

As we know

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

where

$R=109677c{m}^{-1}$

$n_1=2$

$n_2=4$

So putting the values

$\overline{v}=109677(\frac{1}{2^2}-\frac{1}{4^2})c{m}^{-1}$

$=109677(\frac{1}{4}-\frac{1}{16})$

$=20564.4c{m}^{-1}$

we know,

$\overline{v}=\frac{1}{\lambda }$

So $\lambda =\frac{1}{\overline{v}}=\frac{1}{20564.4}$

$=486\times {10}^{-7}cm$

$=486nm.Ans$