(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The position and momentum of a 1.00 KeV electron are simultaneously determined. If the position is located to within 0.100 nm, what is the percentage of uncertainty in its momentum?

(Arthur Beiser - Concepts of modern Physics, 3rd Part exercise 33)

2. Relevant equations

[tex]Δx\cdotΔp≥\frac{\hbar}{2}[/tex]

3. The attempt at a solution

[tex]Δx\cdotΔp≥\frac{\hbar}{2}\rightarrow \; Δp≥ \frac{\hbar}{2\cdotΔx}[/tex]

Given that Δx=0,100 nm, we can evaluate the uncertainty in its momentum Δp.

The momentum of the particle should be of roughly equal magnitude, and evaluating the fraction [tex]\frac{Δp}{p}%[/tex] gives me 98%, where the answer is 3,1 %. What did I do wrong?

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# Homework Help: Heisenberg's uncertainty principle question

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