For a given kinetic energy, which of the following has smaller de Broglie wavelength : electron or proton? Justify your answer.
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The DeBroglie wavelength at non-relativistic velocities is
λ=hp=hmv=hm2Em−−−√=h2Em−−−−√
using
E=12mv2⟺v=2Em−−−√
Since mp>>me, for the same kinetic energy E the DeBroglie wavelength for the proton is much smaller (by the factor memp−−−√) than that of the electron.
So to answer your question: the electron has the larger DeBroglie wavelength. This qualitative answer does not change if you consider relativistic velocities.
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