Question

The de-broglie wavelength of a particle moving with a velocity $2.25\times 10^{8}m/s$ is equal to the wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is:

• A. $3/8$
• B. $5/8$
• C. $7/8$
• D. $5/9$

Answer 1

Answer: (A)

$K_{\text{particle }}=\frac{1}{2}m{v}^2\left[\lambda =\frac{h}{mv}\right]$
$K_{\text{particle }}=\frac{1}{2}\left(\frac{h}{\lambda v}\right)\cdot v^2$
$K_{\text{particle }}=\frac{vh}{2\lambda }$
$K_{\text{photon }}=\frac{hc}{\lambda }$
$\therefore \frac{K_{\text{particle }}}{K_{\text{photon }}}$ $=\frac{v}{2c}$
$=\frac{2.25\times 10^8}{2\times 3\times 10^8}=\frac{3}{8}$