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)

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