Physics - Modern Physics Question with Solution | TestHub

The wave number of the emitted photon when an electron transitions from the fifth energy level ($n=5$) to the ground level ($n=1$) is given by Rydberg's formula:

$\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{5^2}\right)=R\left(1-\frac{1}{25}\right)=\frac{24R}{25}$

The momentum of the emitted photon is:

$p=\frac{h}{\lambda }=h\left(\frac{24R}{25}\right)=\frac{24Rh}{25}$

By the principle of conservation of linear momentum, the momentum acquired by the recoiling Hydrogen atom is equal in magnitude to the momentum of the emitted photon. Thus, the momentum of the atom is:

$p_{\text{atom}}=p=\frac{24Rh}{25}$

The velocity $v$ acquired by the Hydrogen atom of mass $m$ is:

$v=\frac{p_{\text{atom}}}{m}=\frac{24Rh}{25m}$

Answer: $\frac{24\mathrm{Rh}}{25\mathrm{m}}$