If a photon,an electron,and a uranium nucleus | vedclass

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(A) The de Broglie wavelength $\lambda$ is related to momentum $p$ as $\lambda = \frac{h}{p}$.For a particle with mass $m$ and kinetic energy $E$,the

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(A) The de Broglie wavelength $\lambda$ is related to momentum $p$ as $\lambda = \frac{h}{p}$.For a particle with mass $m$ and kinetic energy $E$,the momentum is $p = \sqrt{2mE}$,so $\lambda = \frac{h}{\sqrt{2mE}}$.Rearranging for energy,we get $E = \frac{h^2}{2m\lambda^2}$.For a photon,the energy is $E = \frac{hc}{\lambda}$.Comparing the two,since the mass $m$ of the uranium nucleus is much larger than the mass of the electron,and the photon's energy expression is inversely proportional to $\lambda$ (rather than $\lambda^2$),the photon has the highest energy for a given wavelength $\lambda$.

If a photon,an electron,and a uranium nucleus have the same de Broglie wavelength,which one will have the highest energy?

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