An electron jumps from 5^th orbit to 4^th orbit of hydrogen atom. Taking the Rydberg constant as 10⁷ per metre. What will be the frequency of radiation emitted.

$6.75 \times 10^{13}$ Hz

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