Question

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. $\frac{P_e}{P_h}=\frac{2c}{v}$

Answer 1

The correct option is A

$\frac{E_e}{E_p}=\frac{v}{2c}$

As,
$E_e=\frac{1}{2}m{v}^2=\frac{1}{2}\left(mv\right)v=\frac{1}{2}\left(\frac{h}{\lambda }\right)v$
and $E_p=\frac{hc}{\lambda }$
$\therefore \frac{E_e}{E_p}=\frac{v}{2c}$
Now, $P_e=mv=\frac{h}{\lambda }$
and $P_h=\frac{h}{\lambda }$
$\therefore \frac{P_e}{P_h}=1$