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Question
Suppose that the uncertainty in determining the position of an electron in an orbit is 0.6 Å. What is the uncertainty in its momentum?
Solution
Heisenberg Uncertainity Principle is `∆"x".∆"p" ≥ "h"/(4π)`
Given: ∆x = 0.6 Å = 0.6 × 10–10 m
h = 6.626 × 10–34 kgm2s–1
Uncertainity in momentum,
`∆"p" ≥ "h"/(4π∆"x")`
∆p ≥ `(6.626 xx 10^-34 "kgm"^2 "s"^-1)/(4 xx 3.14 xx 0.6 xx 10^-10 "m")`
∆p ≥ 8.8 × 10–25 kgms–1
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