Question

The de-Broglie wavelength of an electron moving with a velocity $1.5\times 10^{8}m/s$ is equal to that of a photon. The ratio of the kinetic energy of the electron to the energy of the photon is

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

Answer 1

Answer: (A)

$K_e=\frac{1}{2}m{v}^2$ and
$\lambda =\frac{h}{mv}{K}_e=\frac{1}{2}\left(\frac{h}{\lambda v}\right)\cdot v^2$ $=\frac{vh}{2\lambda }$
$K_p=\frac{hc}{\lambda }$
$\therefore \frac{K_e}{K_p}=\frac{v}{2c}=\frac{1.5\times 10^8}{2\times 3\times 10^8}$
$=\frac{1}{4}$