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Show that the de-Broglie wavelength A of an electron of energy E is given by the relation : lambda = h/sqrt(2mE). |
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Answer» SOLUTION :Let E be energy and p MOMENTUM of electron, So `E = 1/2mv^(2) = (m^(2)v^(2))/(2m) =p^(2)/(2m)` or `p=sqrt(2mE)`……(i) Since de-Broglie wavelength associated by an electron is GIVEN by: `lambda = h/p` USING Eq. (i), we get `lambda = h/sqrt(2mE)`...........(ii) |
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