If the de Broglie wavelengths of an electron | vedclass

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.From this,we can express kinetic energy as $E = \frac{h^2}{2m

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Greater than the kinetic energy of the proton

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(D) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.From this,we can express kinetic energy as $E = \frac{h^2}{2m\lambda^2}$.Since $h$ and $\lambda$ are constants for both particles,the kinetic energy $E$ is inversely proportional to the mass $m$,i.e.,$E \propto \frac{1}{m}$.We know that the mass of an electron $(m_e)$ is much smaller than the mass of a proton $(m_p)$,i.e.,$m_e Therefore,since $E$ is inversely proportional to $m$,it follows that $E_e > E_p$.Thus,the kinetic energy of the electron is greater than the kinetic energy of the proton.

If the de Broglie wavelengths of an electron and a proton are equal,then the kinetic energy of the electron is ........

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