The de-Broglie wavelength of a particle movin | vedclass

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(B) Given,the de-Broglie wavelength of the particle $\lambda_p = \frac{h}{mv}$ is equal to the wavelength of the photon $\lambda_{ph} = \frac{h}{p_{ph

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(B) Given,the de-Broglie wavelength of the particle $\lambda_p = \frac{h}{mv}$ is equal to the wavelength of the photon $\lambda_{ph} = \frac{h}{p_{ph}} = \frac{h}{E/c} = \frac{hc}{E_{ph}}$.Since $\lambda_p = \lambda_{ph}$,we have $\frac{h}{mv} = \frac{hc}{E_{ph}}$.Thus,the energy of the photon is $E_{ph} = mvc$.The kinetic energy of the particle is $K_p = \frac{1}{2}mv^2$.The ratio of the kinetic energy of the particle to the kinetic energy of the photon is $\frac{K_p}{E_{ph}} = \frac{\frac{1}{2}mv^2}{mvc} = \frac{v}{2c}$.Substituting the given values $v = 2.25 	imes 10^8 \ m/s$ 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 of $2.25 \times 10^8 \ m/s$ is equal to the wavelength of a photon. The ratio of the kinetic energy of the particle to the kinetic energy of the photon is: (in $/8$)

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