# An electron moving

Question:

An electron moving with speed $\mathrm{v}$ and a photon moving with speed c, have same D-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is:

1. $\frac{3 c}{v}$

2. $\frac{\mathrm{v}}{3 \mathrm{c}}$

3. $\frac{v}{2 c}$

4. $\frac{2 \mathrm{c}}{\mathrm{v}}$

Correct Option: , 3

Solution:

$\lambda_{\mathrm{e}}=\lambda_{\mathrm{Ph}}$

$\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{ph}}}$

$\sqrt{2 \mathrm{mk}_{\mathrm{e}}}=\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{c}}$

$2 \mathrm{mk}_{\mathrm{e}}=\frac{\left(\mathrm{E}_{\mathrm{ph}}\right)^{2}}{\mathrm{c}^{2}}$

$\frac{\mathrm{k}_{\mathrm{e}}}{\mathrm{E}_{\mathrm{ph}}}=\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{c}^{2}}\left(\frac{1}{2 \mathrm{~m}}\right)$

$=\frac{\mathrm{p}_{\mathrm{ph}}}{\mathrm{c}}\left(\frac{1}{2 \mathrm{~m}}\right)$

$=\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{c}}\left(\frac{1}{2 \mathrm{~m}}\right)$

$=\frac{\mathrm{mv}}{\mathrm{c}} \frac{1}{2 \mathrm{~m}}$

$=\frac{\mathrm{v}}{2 \mathrm{c}}$