Question

The wavelength of a photon and de-Broglie wavelength an electron have the same value. Given that $v$ is the speed of electron and $c$ is the velocity of light. $E_{e}, \, E_{p}$ is the kinetic energy of electron and energy of photon respectively while $p_{e}, \, p_{h}$ is the momentum of electron and photon respectively. Then which of the following relation is correct?

• A. $\frac{E_{e}}{E_{p}}=\frac{v}{2 c}$
• B. $\frac{E_{e}}{E_{p}}=\frac{2 c}{v}$
• C. $\frac{p_{e}}{p_{h}}=\frac{c}{2 v}$
• D. $\frac{p_{e}}{p_{h}}=\frac{2 c}{v}$

Answer 1

Answer: (A)

$E_{e}=\frac{1}{2}mv^{2}=\frac{1}{2}\left(m v\right)v=\frac{1}{2}\left(\frac{h}{\lambda }\right)v$
and $E_{p}=\frac{h c}{\lambda };$
$\therefore \, \frac{E_{e}}{E_{p}}=\frac{v}{2 c}$
and according to De Broglie's equation
$p_{h}=\frac{h}{\lambda }$
$\therefore \, \frac{p_{e}}{p_{h}}=1$