What can be said about the de Broglie wavelen | vedclass

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(A) 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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$\lambda_p < \lambda_e$

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(A) 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 kinetic energy $E$ is the same for both the electron and the proton,we have $\lambda \propto \frac{1}{\sqrt{m}}$.We know that the mass of a proton $(m_p)$ is much greater than the mass of an electron $(m_e)$,i.e.,$m_p > m_e$.Therefore,$\sqrt{m_p} > \sqrt{m_e}$,which implies $\frac{1}{\sqrt{m_p}} Thus,the de Broglie wavelength of the proton is less than that of the electron,i.e.,$\lambda_p < \lambda_e$.

What can be said about the de Broglie wavelengths of an electron and a proton having the same kinetic energy?

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