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Show that the de broglie wavelength of an electron that has been accelerated from rest through a potential difference v is given by: Λ = h 2 m e v \\lambda = \\frac { h } { \\sqrt { 2 m e v } } λ = 2 m e v h. Calculate the ratio of the de broglie wavelength of a proton to that of an alpha particle when both have been accelerated from rest by the same potential difference. Where ek = kinetic energy of the electron in electron volt, v = potential difference of the electron, e = charge of electron.