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What is the (a) momentum (b) speed and (c) de-Broglie wavelength of an electron with kinetic energy of 120 eV. Given `h=6.6xx10^(-34)Js, m_(e)=9xx10^(-31)kg , 1eV=1.6xx10^(-19)J`. |
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Answer» (a) `p=sqrt(2mK)=sqrt(2xx(9xx10^(-31))xx(120xx1.6xx10^(-19)))=5.88xx10^(-24)kgms^(-1)` (b) `v=sqrt((2K)/m)=sqrt((2xx(120xx1.6xx10^(-19)))/(9xx10^(-31)))=6.53xx10^(6)ms^(-1)` (c) `lambda=h/p=(6.63xx10^(-34))/(5.88xx10^(-24))=1.13xx10^(-10)m=1.13Å` |
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