Q. An electron jumps from the $4^{th}$ orbit to the $2^{nd}$ orbit of hydrogen atom. Given the Rydberg’s constant $R = 10^{5}cm^{- 1}.$ The frequency in $Hz$ of the emitted radiation will be
A) $\frac{3}{16} \times 10^{5}$
B) $\frac{3}{16} \times 10^{15}$
C) $\frac{9}{16} \times 10^{15}$
D) $\frac{3}{4} \times 10^{15}$
Explanation
(c) $\frac{1}{\lambda} = R\left( \frac{1}{2^{2}} - \frac{1}{4^{2}} \right) = \frac{3R}{16} \Rightarrow \lambda = \frac{16}{3R} = \frac{16}{3} \times 10^{- 5}cm$
Frequency $n = \frac{c}{\lambda} = \frac{3 \times 10^{10}}{\frac{16}{3} \times 10^{- 5}} = \frac{9}{16} \times 10^{15}Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.