Question

The de Broglie wavelength of an electron moving with a velocity $1.5\times {10}^8{ms}^{-1}$ is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the energy of photon is (apply non relativistic formula for electron) :

• A. $2$
• B. $4$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D $\frac{1}{4}$

De Broglie wavelength,

${\lambda }_e=\frac{h}{P_e}=\frac{h}{m{V}_e}{\lambda }_e={\lambda }_p\to \frac{h}{m{V}_e}=\frac{hc}{E}\to \frac{m{V}_e}{E}=\frac{1}{c}\to \frac{1}{2}m{v}^2\times \frac{2}{V_e}=\frac{1}{c}\to \frac{K{E}_e}{E}=\frac{V_e}{2c}=\frac{1.5\times {10}^8}{2\times 3\times {10}^8}\frac{K{E}_e}{E}=\frac{1}{4}$