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

Explanation

Here, `Deltax=1nm=10^(-9)m, Deltap= ?`
As, `DeltaxDeltap~~h :. Deltap=h/(Deltax)=h/(2piDeltax) =(6.62xx10^(-34)Js)/(2xx(22//7)10^(-9)m)=1.05xx10^(-25)kgms^(-1)`
Energy, `E=(p^(2))/(2m)=((Deltap)^(2))/(2m)=((1.05xx10^(-25))^(2))/(2xx9.1xx10^(-31))J = ((1.05xx10^(-25))^(2))/(2xx9.1xx10^(-31)xx1.6xx10^(-19)) eV=3.8xx10^(-2)eV`
(Given `p= Deltap`)