Question

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

• A. $\frac{16}{3R}$
• B. $\frac{2R}{16}$
• C. $\frac{3R}{16}$
• D. $\frac{4R}{16}$

Answer 1

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

Given $n_1=2,n_2=4,$

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

$=R[\frac{1}{4}-\frac{1}{16}]=\frac{3R}{16}$