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

Options

A. `1.506xx10^(4)`
B. `3xx10^(4)`
C. `6xx10^(4)`
D. `9xx10^(4)`

Answer

A

Explanation

`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`