Q. 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 (velocity of light is $3 \times 10^{8}m/s$)
A) 1/8
B) 3/8
C) 5/8
D) 7/8
Explanation
(b) $K_{particle} = \frac{1}{2}mv^{2}\ also\ \lambda = \frac{h}{mv}$
$\Rightarrow K_{particle} = \frac{1}{2}\left( \frac{h}{\lambda v} \right).v^{2} = \frac{vh}{2\lambda}\ \ \ \ \ldots(i)$
$K_{photon} = \frac{hc}{\lambda}\ \ \ \ \ \ \ \ldots(ii)$
$\therefore\frac{K_{particle}}{K_{photon}} = \frac{v}{2c} = \frac{2.25 \times 10^{8}}{2 \times 3 \times 10^{8}} = \frac{3}{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.