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Calculate the a. momentum, and b. de Broglie wavelength of the electrons acceleratedthrough a potential difference of 56V.

Answer»

SOLUTION :a. `v=56V`
`p=mv=sqrt(2mK_("max"))=sqrt(2meV)=sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xx56)`
`=40.38xx10^(-25)KG ms^(-1)=4.038xx10^(-24) kg ms^(-1)`
b. `lambda=(h)/(p)=(6.6xx10^(-34))/(4.038xx10^(-24))=0.163nm`


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