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.

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.



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