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26 June, 21:22

An electron in an atom has an uncertainty of 0.2 nm. If it is doubled to 0.4 nm by what factor does the uncertainty in momentum change?

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  1. 26 June, 21:32
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    The uncertainty in momentum changes by a factor of 1/2.

    Explanation:

    By Heisenberg's uncertainty principle, ΔpΔx ≥ h/2π where Δp = uncertainty in momentum and Δx = uncertainty in position = 0.2 nm. The uncertainty in momentum is thus Δp ≥ h/2πΔx. If the uncertainty in position is doubled, that is Δx₁ = 2Δx = 0.4 nm, the uncertainty in momentum Δp₁ now becomes Δp₁ ≥ h/2πΔx₁ = h/2π (2Δx) = (h/2πΔx) / 2 = Δp/2.

    So, the uncertainty in momentum changes by a factor of 1/2.
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