Question

If the velocity of electron is 25% of the velocity of photon, then $E_e/E_{ph}$ equals

• A. 1:2
• B. 1:4
• C. 1:8
• D. 1:16

Answer 1

The correct option is C 1:8

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)

When velocity of electron is $25$ % of velocity of photon, $v=c\times \left(25/100\right)=c/4$

Thus, $\frac{E_e}{E_{ph}}=\left(c/4\right)/2c=1/8$