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(B) The de-Broglie wavelength of the particle is given by $\lambda = \frac{h}{mv}$,where $m$ is the mass and $v$ is the velocity of the particle.From

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$3$

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(B) The de-Broglie wavelength of the particle is given by $\lambda = \frac{h}{mv}$,where $m$ is the mass and $v$ is the velocity of the particle.From this,we can write $mv = \frac{h}{\lambda}$.The kinetic energy of the particle is $K_{	ext{particle}} = \frac{1}{2}mv^2 = \frac{1}{2}(mv)v = \frac{1}{2} \left( \frac{h}{\lambda} ight) v = \frac{hv}{2\lambda}$ ...$(i)$The energy of a photon with wavelength $\lambda$ is $E_{	ext{photon}} = \frac{hc}{\lambda}$ ...$(ii)$Taking the ratio of the kinetic energy of the particle to the energy of the photon:$\frac{K_{	ext{particle}}}{E_{	ext{photon}}} = \frac{hv / 2\lambda}{hc / \lambda} = \frac{v}{2c}$Given $v = 2.25 	imes 10^8 \ m/s$ and $c = 3 	imes 10^8 \ m/s$:Ratio $= \frac{2.25 	imes 10^8}{2 	imes 3 	imes 10^8} = \frac{2.25}{6} = \frac{225}{600} = \frac{3}{8}$.

The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^8 \ m/s$ is equal to the wavelength of a photon. The ratio of the kinetic energy of the particle to the energy of the photon is (velocity of light is $3 \times 10^8 \ m/s$). (in $/8$)

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