Q. An electron jumps from 5$^{th}$ orbit of 4$^{th}$ orbit of hydrogen atom. Taking the Rydberg constant as $10^{7}per\ metre$ what will be the frequency of radiation emitted
A) $6.75 \times 10^{12}Hz$
B) $6.75 \times 10^{14}Hz$
C) $6.75 \times 10^{13}Hz$
D) None of these
Explanation
(c) By using $v = Rc\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right\rbrack$
$\Rightarrow v = 10^{7} \times \left( 3 \times 10^{8} \right)\left\lbrack \frac{1}{4^{2}} - \frac{1}{5^{2}} \right\rbrack = 6.75 \times 10^{13}Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.