Question

The de-Broglie wavelength of a particle moving with a velocity 2.25 × 10⁸ 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 × 10⁸ m/s)

• A. 1/8
• B. 3/8
• C. 5/8
• D. 7/8

Answer 1

Answer: (B)

3/8

Answer 2

$K_{\text{particle}} = \frac{1}{2}mv^{2}$ also $\lambda = \frac{h}{mv}$

$\Rightarrow K_{\text{particle}} = \frac{1}{2}\left( \frac{h}{\lambda v} \right).v^{2} = \frac{vh}{2\lambda}$ ...(i)

$K_{\text{photon}} = \frac{hc}{\lambda}$ ...(ii)

$\therefore\frac{K_{\text{particle}}}{K_{\text{photon}}} = \frac{v}{2c} = \frac{2.25 \times 10^{8}}{2 \times 3 \times 10^{8}} = \frac{3}{8}$