Question

An electron and a photon possess the same de-Broglie wavelength. If $E_e$ and $E_p$ respectively are the energies of electron and photon and v and c are their respective velocities, then $\frac{E_e}{E_p}$ is equal to

• A. $\frac{v}{c}$
• B. $\frac{v}{2c}$
• C. $\frac{v}{3c}$
• D. $\frac{v}{4c}$

Answer 1

Answer: (B)

de-Broglie wavelength $\lambda$
= $\frac{h}{\sqrt{2m{E}_e}}=\frac{hc}{E_p}$ or $2m{E}_e=\frac{E_p^2}{c^2}$ ... ( i)
But $E_e=\frac{1}{2}m{v}^2$ or $m=\frac{2E_e}{v^2}$
From Eq. (i)
2 $\left[\frac{2E_e}{v^2}\right]E_e=\frac{E_p^2}{c^2}$
or $\frac{4E_e^2}{v^2}=\frac{E_p^2}{c^2}$ or $\frac{E_e^2}{E_p^2}=\frac{v^2}{4c^2}$
or $\frac{E_e}{E_p}=\frac{v}{2c}$