An electron and a proton have the same de-Bro | vedclass

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(D) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is the mass of the

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Greater than the kinetic energy of the proton

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(D) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $E$ is the kinetic energy.Since the de-Broglie wavelength $\lambda$ is the same for both the electron and the proton,we have $\lambda = 	ext{constant}$.From the formula,we can see that $E = \frac{h^2}{2m\lambda^2}$,which implies $E \propto \frac{1}{m}$.Since the mass of an electron $(m_e)$ is much smaller than the mass of a proton $(m_p)$,i.e.,$m_e Therefore,$E_e > E_p$.

An electron and a proton have the same de-Broglie wavelength. Then the kinetic energy of the electron is

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