An electron jumps from the 4^th orbit to the 2^nd orbit of hydrogen atom. Given the Rydberg's constant R = 10⁵ cm^-1. The frequency in Hz of the emitted radiation will be :

$\frac{9}{16} \times 10^{15}$

$\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3 R}{16} \Rightarrow \lambda=\frac{16}{3 R}=\frac{16}{3} \times 10^{-5} cm$

Frequency  $n=\frac{c}{\lambda}=\frac{3 \times 10^{10}}{\frac{16}{3} \times 10^{-5}}=\frac{9}{16} \times 10^{15} Hz$