- #1

qban88

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The figure shows 1.0*10^-6 m diameter dust particles in a vacuum chamber. The dust particles are released from rest above a 1.0*10^-6 m diameter hole, fall through the hole (there's just barely room for the particles to go through), and land on a detector at distance d below.

If the particles were purely classical, they would all land in the same 1.0 micrometer diameter circle. But quantum effects don't allow this. If d = 1.3 m , by how much does the diameter of the circle in which most dust particles land exceed 1.0 micrometer ?

PART B) Quantum effects would be noticeable if the detection-circle diameter increased to 1.7 micrometers. At what distance would the detector need to be placed to observe this increase in the diameter?

2. Homework Equations

h/2 <= ΔxΔp

Vf = (2*g*d)^(1/2) for a free fall body starting at V = 0.

p = mv

3. The Attempt at a Solution

If I understand it correctly, the suggestion to this problem posted previously in the forum (https://www.physicsforums.com/showthread.php?t=352999) does not work. I tried using λ=h/p finding p by using the velocity of a free fall body given displacement d and λ would be the scattering but it does not work.

I also tried using the uncertainty principle assuming the uncertainty in Δx is 1.0 micrometers initially and from there I can find Δp and therefore Δv = 3.315*10^-13 but I don't know how to relate this to the distance traveled d in order to find Δx final which is what is being asked. Does the uncertainty in velocity increases as the particle moves and if so, in what fashion? Should I just do this as a particle diffraction problem? Having the formula or idea for part A would yield the solution to part B i believe.

Any help would be appreciated