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What is the (a) momentum, (b) speed, and (c) de-Broglie wavelength of an electro with kinetic energy energy of 120 eV. |
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Answer» Solution :Kinetic energy of electron `K=120eV=120xx1.6xx10^(-19)J=1.92xx10^(-17)J` (a) `THEREFORE `Momentum of electron `p=sqrt(2mK)=sqrt(2xx9.11xx10^(-31)xx1.92xx10^(-17))=5.92xx10^(-24)kg" ms"^(-1)` (b) Speed of electron `v=(p)/(m)=(5.92xx10^(-24))/(9.11xx10^(-31))=6.50xx10^(6)ms^(-1)` (C) de-Broglie wavelength of electron `lamda=(h)/(p)=(6.63xx10^(-34))/(5.92xx10^(-24))=1.12xx10^(-10)m=0.112nm`. |
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