Answer:
de-Broglie wavelength of a particle of mass m and kinetic energy K is \[\lambda =\frac{h}{\sqrt{2mK}}\] or \[{{\lambda }^{2}}=\frac{{{h}^{2}}}{2mK}\] \[\therefore \] \[K=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\] For particles having same\[\lambda ,K\propto \frac{1}{m}\] As the mass of electron is smaller than that of proton, so the electron has a higher kinetic energy and it is moving faster.
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