Physics - Modern Physics Question with Solution | TestHub

$mv=\frac{h}{\lambda }$, here $v$ is recoil speed of $\mathrm{H}$-atom and $\lambda$ is the wavelength of the photon.

When the electron jumps down from the first excited state to the ground state the energy released is

$∆E_{2\to 1}=E\left(1-\frac{1}{2^2}\right)=\frac{3E}{4}$

$\frac{3E}{4}=\frac{1}{2}m{v}^2+\frac{hc}{\lambda }$

$\frac{3E}{4}=\frac{1}{2}m{v}^2+mvc$

 $\left(\frac{m}{2}\right)v^2+\left(mc\right)v-\left(\frac{3E}{4}\right)=0$

$\Rightarrow v=\left[\sqrt{\frac{3E}{2m}+c^2}\right]-c$