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Question
Assuming an electron is confined to a 1 nm wide region, find the uncertainty in momentum using Heisenberg Uncertainty principle (∆x∆p ≃ h). You can assume the uncertainty in position ∆x as 1 nm. Assuming p ≃ ∆p, find the energy of the electron in electron volts.
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Solution
∆x∆p ≃ h
∆p ≃ `h/(∆x) ≃ (1.05 xx 10^-34 Js)/(10^-9 m)` = 1.05 × 10–25
E = `p^2/(2m) = (1.05 xx 10^-25)^2/(2 xx 9.1 xx 10^-31)`
= `1.05^2/18.2 xx 10^-19 J`
= `1.05^2/(18.2 xx 1.6) eV`
= 3.8 × 10–2 eV
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