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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. |
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Answer» Solution :Kinetic ENERGY , `K=120 eV=120xx1.6xx10^(-19)J` `=1.92xx10^(-17)J` `=1.92xx10^(-17)J` `h=6.63xx10^(-34)Js` `m=9.1xx10^(-31)Kg` (a)Momentum of electron, `p=sqrt(2mK)` `therefore p=sqrt(2xx9.1xx10^(-31)xx1.92xx10^(-17))` `therefore p=sqrt(34.944xx10^(-48))` `therefore p=5.91xx10^(-24) kgms^(-1)` (B)Speed of electron, `v=(p)/(m)=(5.91xx10^(-24))/(9.1xx10^(-31))` `therefore v=0.649xx10^(7)` `therefore v~~6.5xx10^(6)m//s` (c )de-Broglie wavelength, `LAMBDA=(h)/(p)` `therefore lambda=(6.63xx10^(-34))/(5.91xx10^(-24))` `therefore lambda=1.121xx10^(-10)m` `therefore lambda=0.1121 nm` |
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