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tatuchka [14]
3 years ago
5

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?
Physics
1 answer:
fenix001 [56]3 years ago
4 0

Answer:

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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To solve this problem it is necessary to simply apply the concepts related to cross-multiply and proportion between units.

Let's start first by relating the amount of dose needed to be supplied per hour, in other words,

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3 years ago
What is transmitted by all waves
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<h3><u>Explanation;</u></h3>
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