Question

A photon has same wavelength as the de Broglie wavelength of electrons. Given $C=$ speed of light, $v=$ speed of electron. Which of the following relation is correct? [Here $E_e=$ kinetic energy of electron, $E_{ph}=$ energy of photon, $P_e=$ momentum of electron and $P_{ph}=$ momentum of photon]

• A. $E_e/E_{ph}=2C/v$
• B. $E_e/E_{ph}=v/2C$
• C. $P_e/P_{ph}=2C/v$
• D. $P_e/P_{ph}=C/v$

Answer 1

The correct option is B $E_e/E_{ph}=v/2C$

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

Now, ${\lambda }_e=\frac{h}{P_e}=\frac{h}{mv}....\left(2\right)$

Energy of photon, $E_{ph}=\frac{hC}{{\lambda }_{ph}}=\frac{hC}{{\lambda }_e}$ (using (1))

Energy of electron, $E_e=\frac{P_e^2}{2m}=\frac{h^2}{{\lambda }_e^2\left(2m\right)}$ using (2)

Now, $\frac{E_e}{E_{ph}}=\frac{h^2}{{\lambda }_e^2\left(2m\right)}\times \frac{{\lambda }_e}{hC}=\left(1/2mC\right)\left(h/{\lambda }_e\right)=\left(1/2mC\right)\left(mv\right)=v/2C$ using (2)

Now, $\frac{P_e}{P_{ph}}=\frac{h/{\lambda }_e}{h/{\lambda }_{ph}}=1$ using (1)