Question

The de-Broglie wavelength of a photon and electron are same. $E_{e}, \, E_{p}$ are the kinetic energy of the electron and photon, respectively. $p_{e}, \, p_{h}$ are the momentum of electron and photon, respectively. Then, which of the following relations are correct?
$\left(\right.v=\text{speed of electron},c=\text{velocity of light}\left.\right)$

• 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 $E_{p}=\frac{h c}{\lambda }$
$\therefore \, \frac{E_{e}}{E_{p}}=\frac{v}{2 c}$
According to de-Broglie's equation,
$p_{h}=\frac{h}{\lambda }$
$\therefore \, \frac{p_{e}}{p_{h}}=1$