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.

Explanation

Here, `Deltax=1nm = 10^(-9)m, Deltap=?`
`AsDeltax Deltap~~h`
`therefore" "Deltap=h/(Deltax)=h/(2piDeltax)`
`implies " "=(6.62xx10^(-34)JS)/(2xx(22//7)(10^(-9))m)`
Energy, `" "E=(p^(2))/(2m)=((Deltap)^(2))/(2m)" "[thereforep~~Deltap]`
`" "=((1.05xx10^(-25)))/(2xx9.1xx10^(-31))J`
`implies" "=((1.05xx10^(-25))^(2))/(2xx9.1xx10^(-31)xx1.6xx10^(-19))eV`
`" "=38xx10^(-2)eV.`