Q. A photon and electron have same de-Broglie wavelength. Give that $v$ is the speed of electron and $c$ is the velocity of light. $E_{e},\ E_{p}$ are the kinetic energy of electron and photon respectively. $p_{e},\ p_{h}$ are the momentum of electron and photon respectively. Then which of the following relation is correct?
A) $\frac{E_{e}}{E_{p}} = \frac{v}{2c}$
B) $\frac{E_{e}}{E_{p}} = \frac{2c}{v}$
C) $\frac{p_{e}}{p_{h}} = \frac{c}{2v}$
D) $\frac{p_{e}}{p_{h}} = \frac{2c}{v}$
Explanation
(a) $E_{e} = \frac{1}{2}mv^{2} = \frac{1}{2}(mv)v = \frac{1}{2}\left( \frac{h}{\lambda} \right)v$
and $E_{p} = \frac{hc}{\lambda};$
$\therefore\ \frac{E_{e}}{E_{p}} = \frac{v}{2c}$
$p_{e} = mv = h/\lambda$ and $p_{h} = \frac{h}{\lambda}$
$\therefore\ \frac{p_{e}}{p_{h}} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.