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(B) Given that the de-Broglie wavelength of the particle is equal to the wavelength of the photon: $\lambda_p = \lambda_{ph} = \lambda$.For a photon,t

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(B) Given that the de-Broglie wavelength of the particle is equal to the wavelength of the photon: $\lambda_p = \lambda_{ph} = \lambda$.For a photon,the energy is $E_{ph} = \frac{hc}{\lambda}$.For a particle,the de-Broglie wavelength is $\lambda = \frac{h}{mv}$,which implies $h = \lambda mv$.The kinetic energy of the particle is $K_p = \frac{1}{2}mv^2$.Substituting $h$ into the photon energy equation: $E_{ph} = \frac{(\lambda mv)c}{\lambda} = mvc$.The ratio of the kinetic energy of the particle to the energy of the photon is:$\frac{K_p}{E_{ph}} = \frac{\frac{1}{2}mv^2}{mvc} = \frac{v}{2c}$.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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