Q. An electron makes a transition from orbit $n = 4$ to the orbit $n = 2$ of a hydrogen atom. The wave number of the emitted radiations ($R =$ Rydberg’s constant) will be
A) $\frac{16}{3R}$
B) $\frac{2R}{16}$
C) $\frac{3R}{16}$
D) $\frac{4R}{16}$
Explanation
(c) Wave number $\frac{1}{\lambda} = R\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack = R\left\lbrack \frac{1}{4} - \frac{1}{16} \right\rbrack = \frac{3R}{16}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.