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(D) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.Since t

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$K_{p} < K_{e}$

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(D) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.Since the proton and electron have the same de Broglie wavelength,their momenta must be equal: $p_p = p_e = p$.The kinetic energy $K$ of a particle is related to its momentum $p$ and mass $m$ by the formula $K = \frac{p^2}{2m}$.For the proton and electron,we have $K_p = \frac{p^2}{2m_p}$ and $K_e = \frac{p^2}{2m_e}$.Since the mass of an electron $m_e$ is much smaller than the mass of a proton $m_p$ $(m_e  \frac{1}{2m_p}$.Therefore,$K_e > K_p$,or $K_p < K_e$.

$A$ proton and an electron have the same de Broglie wavelength. If $K_p$ and $K_e$ are the kinetic energies of the proton and electron respectively,then choose the correct relation:

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