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(B) For an electron,the de Broglie wavelength is given by $\lambda = \frac{h}{p}$,which implies $p = \frac{h}{\lambda}$.For a photon,the energy is giv

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$3.33 	imes 10^{-9}$

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(B) For an electron,the de Broglie wavelength is given by $\lambda = \frac{h}{p}$,which implies $p = \frac{h}{\lambda}$.For a photon,the energy is given by $E = \frac{hc}{\lambda}$.We are asked to find the ratio $p/E$.Substituting the expressions for $p$ and $E$:$\frac{p}{E} = \frac{h/\lambda}{hc/\lambda} = \frac{h}{\lambda} 	imes \frac{\lambda}{hc} = \frac{1}{c}$.Given the speed of light $c = 3 	imes 10^8 \ m/s$,we have:$\frac{p}{E} = \frac{1}{3 	imes 10^8} = 0.333 	imes 10^{-8} = 3.33 	imes 10^{-9} \ s/m$.

The wavelength of an electron and a photon are equal. If the energy of the photon is $E$ and the momentum of the electron is $p$,what is the value of $p/E$?

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