A photon and an electron have equal energy E. | vedclass

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(B) For a photon,the energy is given by $E = \frac{hc}{\lambda_{	ext{photon}}}$,which implies $\lambda_{	ext{photon}} = \frac{hc}{E}$.For an electro

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$1/\sqrt{E}$

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(B) For a photon,the energy is given by $E = \frac{hc}{\lambda_{	ext{photon}}}$,which implies $\lambda_{	ext{photon}} = \frac{hc}{E}$.For an electron,the de Broglie wavelength is given by $\lambda_{	ext{electron}} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,where $m$ is the mass of the electron and $E$ is its kinetic energy.Taking the ratio of the two wavelengths:$\frac{\lambda_{	ext{photon}}}{\lambda_{	ext{electron}}} = \frac{hc/E}{h/\sqrt{2mE}} = \frac{hc}{E} \cdot \frac{\sqrt{2mE}}{h} = c \sqrt{\frac{2m}{E}}$.Since $c$ and $m$ are constants,we have $\frac{\lambda_{	ext{photon}}}{\lambda_{	ext{electron}}} \propto \frac{1}{\sqrt{E}}$.

$A$ photon and an electron have equal energy $E$. The ratio $\lambda_{\text{photon}} / \lambda_{\text{electron}}$ is proportional to:

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