1.

The recoil energy of an electron of wavelength 0.1Å in eV will be :

Answer»

`1.506xx10^(4)`
`3XX10^(4)`
`6XX10^(4)`
`9xx10^(4)`

Solution :`E_(r)=(p^(2))/(2m),p(h)/(lambda) :. E_(r)=(h^(2))/(2m lambda^(2)) K`
`E_(r)=(h^(2))/(1*6xx10^(-19)2 m lambda^(2)) eV`
`=(6*62xx10^(-34))/(2xx9*1xx10^(-31)xx(0*1xx10^(-10))^(2)XX1*6xx10^(-19))`
`E_(r)=1*506xx10^(4)eV`


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