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(C) $1$. Energy of a photon $(E_p)$ is given by $E_p = \frac{hc}{\lambda}$.$2$. De-Broglie wavelength of an electron is $\lambda = \frac{h}{p}$,where

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$\frac{h}{2 \lambda m c}$

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(C) $1$. Energy of a photon $(E_p)$ is given by $E_p = \frac{hc}{\lambda}$.$2$. De-Broglie wavelength of an electron is $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron. Thus,$p = \frac{h}{\lambda}$.$3$. Kinetic energy of the electron $(K_e)$ is given by $K_e = \frac{p^2}{2m}$.$4$. Substituting $p = \frac{h}{\lambda}$ into the kinetic energy formula: $K_e = \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2}$.$5$. The ratio of the kinetic energy of the electron to the energy of the photon is $\frac{K_e}{E_p} = \frac{h^2 / (2m\lambda^2)}{hc / \lambda}$.$6$. Simplifying the expression: $\frac{K_e}{E_p} = \frac{h^2}{2m\lambda^2} 	imes \frac{\lambda}{hc} = \frac{h}{2mc\lambda}$.

The wavelength $\lambda$ of a photon and the de-Broglie wavelength of an electron have the same value. The ratio of the kinetic energy of the electron to the energy of a photon is ($m=$ mass of electron,$c=$ velocity of light,$h=$ Planck's constant).

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