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(C) For a photon,the energy is given by $E = hf = \frac{hc}{\lambda}$.For an electron,the momentum is given by $p = \frac{h}{\lambda}$.Taking the rati

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$3.00 	imes 10^8$

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(C) For a photon,the energy is given by $E = hf = \frac{hc}{\lambda}$.For an electron,the momentum is given by $p = \frac{h}{\lambda}$.Taking the ratio of energy of the photon to the momentum of the electron:$\frac{E}{p} = \frac{hc / \lambda}{h / \lambda} = c$.Given that the wavelength $\lambda$ is the same for both the photon and the electron,the ratio depends only on the speed of light $c$.Substituting the value of $c = 3 	imes 10^8 \, m/s$,we get:$\frac{E}{p} = 3 	imes 10^8 \, m/s$.

An electron and a photon have the same wavelength of $10^{-9} \, m$. If $E$ is the energy of the photon and $p$ is the momentum of the electron,then the magnitude of $E / p$ (in $SI$ unit) is

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