An electron of stationary Hydrogen atom passes from fifth energy level to ground level. The velocity that the atom acquired as a result of photo emission is
( \(\mathrm{m}=\) mass of electron, \(\mathrm{R}=\) Rydberg's constant)
( \(\mathrm{h}=\) Planck's constant)

\(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2{ }^2}\right)\)
In this case, \(\mathrm{n}_1=1\) and \(\mathrm{n}_2=5\)
\(\therefore \quad \frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{5^2}\right)=\frac{24}{25} R...(i)\)
Momentum of Photon,
\(\mathrm{p}=\frac{\mathrm{h}}{\lambda}=\mathrm{h}\left(\frac{24}{25} \mathrm{R}\right)\) ...[From (i)]
By conservation of momentum, Momentum of Photon \(=\) Momentum of atom
\(\begin{aligned}
& \quad \mathrm{h}\left(\frac{24}{25} \mathrm{R}\right)=\mathrm{mv} \\
& \therefore \quad \mathrm{v}=\frac{24 \mathrm{Rh}}{25 \mathrm{~m}}
\end{aligned}\)