The de Broglie wavelength of an electron moving with a velocity of 1.5 x 10⁸ m/ sec is equal to that of a photon. Find the ratio of the energy of the photon to that of the kinetic energy of the electron.

Speed of photon (c) = 3 x 10⁸ m/ sec . Let $\mathrm{\lambda }$ be the wavelength of the photon. The de Broglie wavelength of the electron $\frac{\mathrm{h}}{\mathrm{mv}}$. Given $\mathrm{\lambda }$ = $\frac{\mathrm{h}}{\mathrm{mv}}$, 
Now, 
 $\begin{array}{l}\frac{\text{ energy of photon }}{\text{ K.E. of electron }}=\frac{\mathrm{hv}}{\frac{1}{2}{\mathrm{mv}}^2}=\frac{2\mathrm{hc}}{{\mathrm{mv}}^2\mathrm{\lambda }}\\ =\frac{2\mathrm{c}}{\mathrm{v}}\left(\therefore \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mN}}\mathrm{\&}\mathrm{v}=\frac{\mathrm{c}}{\mathrm{\lambda }}\right)=\frac{2\times 3\times {10}^8}{1.5\times {10}^8}=4\end{array}$