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)

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Aicharya · Accepted Answer

$\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2{ }^2}\right)$
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 $\begin{aligned} & \quad \mathrm{h}\left(\frac{24}{25} \mathrm{R}\right)=\mathrm{mv} \\ & \therefore \quad \mathrm{v}=\frac{24 \mathrm{Rh}}{25 \mathrm{~m}} \end{aligned}$

An electron of stationary Hydrogen atom passes from fifth energy level to ground level. The velocity that…

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