Minimum uncertainty of the momentum of a small particle

AI Thread Summary
To find the minimum uncertainty of momentum for a particle with mass m=1g confined in a region of width a=1cm, the uncertainty principle formula ΔpΔx ≥ ħ/2 is applied. By substituting Δx with 10^(-2) m, the calculation leads to Δp = 10^2 * ħ/2. The mass provided in the problem may be irrelevant to the calculation, suggesting a focus on velocity uncertainty instead. The discussion references additional resources for further understanding of particle confinement and uncertainty principles. The key takeaway is the relationship between position and momentum uncertainties in quantum mechanics.
AndrejN96
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Homework Statement


Find the minimum uncertainty of the momentum of a small particle with mass m=1g, which is confined within a region of width a=1cm.

Homework Equations


Delta(p)*Delta(x)>=hbar/2

The Attempt at a Solution



Delta(p)*Delta(x)=hbar/2
Delta(p)*10^(-2)=hbar/2
Delta(p)=10^2*hbar/2

This looks pretty straightforward to me, but the given mass in this problem is what confuses me.
 
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I think you have it right & putting in the mass is a red herring. Maybe they meant to ask for the minimum uncertainty in the velocity.
 
You might want to scan this:

http://en.wikipedia.org/wiki/Particle_in_a_box

Pull quote:

"The uncertainties in position and momentum (
b56546a86ab832a9b2a5b15f96519319.png
and
7aa41487a1a40b0077afa0c3331ba111.png
) are defined as being equal to the square root of their respective variances, so that:

a83e98aee94cda7f0410e16698b54ebf.png

This product increases with increasing n, having a minimum value for n=1. The value of this product for n=1 is about equal to 0.568
9dfd055ef1683b053f1b5bf9ed6dbbb4.png
..."
 
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