An electron jumps from the \(4^{\text {th }}\) orbit to the \(2^{\text {nd }}\) orbit of hydrogen atoms. Given the Rydberg's constant \(\mathrm{R}_{\mathrm{H}}=10^7 \mathrm{~m}^{-1}\). frequency in \(\mathrm{Hz}\) of the emitted radiation is \(\left(\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right.\) )

Concept: The wavelength \(\lambda\) of electromagnetic radiation emitted in vacuum, then
\(\frac{1}{\lambda}=\text R_\text H\left(\frac{1}{\text n_1^2}-\frac{1}{\text n_2^2}\right)\qquad\ldots(1)\)
The frequency of the emitted radiation is:
\(\text f=\frac{\text c}{\lambda}\qquad\ldots(2)\)
On plugging equation (1) in equation (2),
\(\mathrm{f}=\mathrm{cR}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
Given, \(\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, \mathrm{R}_{\mathrm{H}}=10^7 \mathrm{~m}^{-1}, \mathrm{n}_1=2\) and \(\mathrm{n}_2=4\).
\(\mathrm{f}=3 \times 10^8 \mathrm{~ms}^{-1} \times 10^7 \mathrm{~m}^{-1}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)\)
On solving,
\(\mathrm{f}=\frac{9}{16} \times 10^{15} \mathrm{~Hz}\)