An electron jumps from 5th orbit to 4th orbit of hydrogen atom. Taking the Rydberg constant as 107 per metre. What will be the frequency of radiation emitted.
Answer & explanation
Correct answer: option 3
By using $v=R C\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
$\Rightarrow v=10^7 \times\left(3 \times 10^8\right)\left[\frac{1}{4^2}-\frac{1}{5^2}\right]=6.75 \times 10^{13}$ Hz