A photon and an electron (mass $m$ ) have the same energy $E$. The ratio ( $\lambda_{ ext {photon }

Question

A photon and an electron (mass $m$ ) have the same energy $E$. The ratio ( $\lambda_{	ext {photon }} / \lambda_{	ext {electron }}$ ) of their de-Broglie wavelengths is ( $c$ is the speed of light)

Professor PC Thomas Classes · Answer

$ \begin{aligned} & 	ext { (a) }: \lambda_{p h}=\frac{h c}{E} \ & \lambda_e=\frac{h}{p}=\frac{h}{\sqrt{2 m E}} \end{aligned} $ \ From eqn. (i) and (ii), we get \ $ 	herefore \quad \frac{\lambda_{p h}}{\lambda_e}=\frac{h c \sqrt{2 m E}}{E \cdot h}=c \sqrt{\frac{2 m}{E}} $