Physics - Modern Physics Question with Solution | TestHub

Energy of the photon is given by $E_p=\frac{hc}{\lambda }$

For the electron, the de-Broglie wavelength is $\lambda =\frac{h}{p}\Rightarrow p=\frac{h}{\lambda }$

Kinetic energy of the electron is $E_e=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }^2}$

The ratio of the kinetic energy of the electron to the energy of the photon is:

$\frac{E_e}{E_p}=\frac{\frac{h^2}{2m{\lambda }^2}}{\frac{hc}{\lambda }}$

$\frac{E_e}{E_p}=\frac{h}{2mc\lambda }$

Answer: $\frac{h}{2\lambda mc}$