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(C) Let the wavelength of the photon be $\lambda_p$ and the particle be $\lambda_m$. Given $\lambda_p = \lambda_m = \lambda$.The energy of the photon

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Greater than the kinetic energy of the particle

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(C) Let the wavelength of the photon be $\lambda_p$ and the particle be $\lambda_m$. Given $\lambda_p = \lambda_m = \lambda$.The energy of the photon is $E_p = \frac{hc}{\lambda}$.The kinetic energy of the particle is $K_m = \frac{p^2}{2m}$. Since $p = \frac{h}{\lambda}$,we have $K_m = \frac{h^2}{2m\lambda^2}$.Comparing the two energies,the ratio is $\frac{E_p}{K_m} = \frac{hc/\lambda}{h^2/(2m\lambda^2)} = \frac{2mc\lambda}{h}$.Since $c = 
u\lambda$ and $p = \frac{h}{\lambda} = mv$,we have $\lambda = \frac{h}{mv}$. Substituting this,$\frac{E_p}{K_m} = \frac{2mc(h/mv)}{h} = \frac{2c}{v}$.Since the speed of light $c$ is much greater than the speed of any particle $v$ $(c > v)$,the ratio $\frac{2c}{v} > 1$.Therefore,$E_p > K_m$. The energy of the photon is greater than the kinetic energy of the particle.

If the de Broglie wavelength of a particle is equal to the wavelength of a photon,then the energy of the photon is .....

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