Assuming an electron is confined to a `1nm` wide region, find the wavelength in momentum using Heisenberg Uncertainty principal `(Deltax Deltap~~h)`. You can assume the uncertainty in position `Deltax` and `1nm`. Assuming `p~=Deltap`, find the energy of the electron in electron volts.

Options

A. `1.6 meV`
B. `3.8 meV`
C. `0.16 meV`
D. `0.38 meV`

Answer

D

Explanation

Here, `Deltax=Inm=10^(-9)m,Deltap=?`
As `DeltaxDeltap ~~h`
`thereforep=h/(Deltax)=h/(2piDeltax)`
`=(6.62xx10^(-34)Js)/(2xx(22//7)(10^(-9)))m`
`=1.05xx10^(-25) kg m//s`
Energy,`E=p^(2)/(2m)=(Deltap)^(2)/(2m) [becauserho~~ Deltap]`
`=(1.05xx10^(-25))^(2)/(2xx9.1xx10^(-31))J`
`=(1.05xx10^(-25))^(2)/(2xx9.1xx10^(-31)xx1.6xx10^(-19))eV=3.8xx10^(-2) eV`