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(C) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $KE$ by the formula: $\lambda = \frac{h}{\sqrt{2mKE}}$.Rearranging this for k

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more than that of a proton

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(C) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $KE$ by the formula: $\lambda = \frac{h}{\sqrt{2mKE}}$.Rearranging this for kinetic energy,we get: $KE = \frac{h^2}{2m\lambda^2}$.Since the de-Broglie wavelength $\lambda$ and Planck's constant $h$ are the same for both particles,we have $KE \propto \frac{1}{m}$.Because the mass of an electron $(m_e \approx 9.11 	imes 10^{-31} \ kg)$ is much smaller than the mass of a proton $(m_p \approx 1.67 	imes 10^{-27} \ kg)$,the kinetic energy of the electron will be greater than that of the proton.

If an electron and a proton have the same de-Broglie wavelength,then the kinetic energy of the electron is

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