Question

An electron and a photon possess the same de Broglie wavelength. If $E_e$ and $E_{ph}$ are, respectively, the energies of electron and photon while $v$ and $c$ are their respective velocities, then $E_e/E_{ph}$ is equal to :

• A. $v/c$
• B. $v/2c$
• C. $v/3c$
• D. $v/4c$

Answer 1

The correct option is B $v/2c$

Given, ${\lambda }_e={\lambda }_{ph}...\left(1\right)$

de Broglie wavelength of electron is ${\lambda }_e=\frac{h}{p_e}=\frac{h}{mv}$ $...\left(2\right)$

The energy of electron is $E_e=\frac{p_e^2}{2m}=\frac{(h/{\lambda }_e{)}^2}{2m}$ using (2)

and the energy of photon is $E_{ph}=\frac{hc}{{\lambda }_{ph}}=\frac{hc}{{\lambda }_e}$ using (1)

Thus, $\frac{E_e}{E_{ph}}=\frac{(h/{\lambda }_e{)}^2}{2m}\times \frac{{\lambda }_e}{hc}=\left(\frac{1}{2mc}\right)\frac{h}{{\lambda }_e}=\left(1/2mc\right)\left(mv\right)=v/2c$ using (2)