Assuming an electron is confined to a 1nm wide region, find the uncertainty in momentum using Heisenberg Uncertainty principle (Ref Eq 11.12 of NCERT Textbook). You can assume the uncertainty in position Δx as 1nm. Assuming p ≃ Δp, find the energy of the electron in electron volts.

Question

Assuming an electron is confined to a 1nm wide region, find the uncertainty in momentum using Heisenberg Uncertainty principle (Ref Eq 11.12 of NCERT Textbook). You can assume the uncertainty in position Δx as 1nm. Assuming p ≃ Δp, find the energy of the electron in electron volts.

Philoid · Accepted Answer

According to the Heisenberg’s Uncertainty Principle

Here, h is the Plank’s constant, ∆x is the uncertainty in the position and ∆p is the uncertainty in the momentum of the electron. ∆x is given and is equal to 1nm = 10^-9 m

As given in the question, we can treat ∆p as p in order to find the energy of the electron

Converting to eV by dividing by 1.6 x 10^-19.