InterviewSolution
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InterviewSolution
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Calculate the (a) momentum and (b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56V. Given, `h=6.63xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg, e=1.6xx10^(-19)C`. |
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Answer» (a) `eV=1/2mv^(2) or mv = sqrt(2eVm)` Momentum of electron, `mv=sqrt(2xx(1.6xx10^(-19))xx56xx9xx10^(-31))=4.02xx10^(-24)kgms^(-1)` (b) `lambda=12.27/(sqrtV)Å=12.27/(sqrt(56))=1.64Å` |
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