Question

An electron makes a transition from orbit $n=4$ to the orbit $n=2$, of a hydrogen atom. The wavenumber of emitted radiation will be,
(Consider $R=$ Rydberg's constant )

• A. $\frac{16}{3R}$
• B. $\frac{R}{8}$
• C. $\frac{3R}{16}$
• D. $\frac{R}{4}$

Answer 1

The correct option is C $\frac{3R}{16}$

The wave number $\left(\frac{1}{\lambda }\right)$ is given by,
$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{1}{\lambda }=R[\frac{1}{2^2}-\frac{1}{4^2}]$

$\therefore \frac{1}{\lambda }=\frac{3R}{16}$

Hence, $\left(C\right)$ is the correct answer.

Why this Question? This question is based on wavenumber which is the reciprocal of the wavelength of emitted radiation.