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

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The photon

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.For a particle with mass $m$ and kinetic energy $E$,the momentum is $p = \sqrt{2mE}$. Thus,$\lambda = \frac{h}{\sqrt{2mE}}$.Rearranging for energy,we get $E = \frac{h^2}{2m\lambda^2}$.Since the wavelength $\lambda$ is the same for all,the energy $E$ is inversely proportional to the mass $m$ $(E \propto \frac{1}{m})$.For a photon,the energy is given by $E = \frac{hc}{\lambda}$. Comparing this to the massive particles,the photon behaves as if it has the smallest effective mass in this context,leading to the highest energy for a given wavelength.Therefore,the photon has the most energy.

$A$ photon,an electron,and a uranium nucleus all have the same wavelength. The one with the most energy is:

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