Heisenberg Uncertainty Principle or Diffraction problem

[qban88]

Homework Statement

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?

Homework Equations

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

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

 

[anathema]

.

First of all, it is important to clarify the assumptions and limitations of this problem. Since the particles are being released from rest and are falling through a small hole, we can assume that their initial velocity is zero and their final velocity is equal to the velocity they acquire during free fall (assuming no air resistance). Therefore, we can use the equation Vf = (2*g*d)^(1/2) to calculate the final velocity of the particles.

Now, let's look at the uncertainty principle. The uncertainty in position, Δx, is given by the diameter of the circle in which most particles will land, which is what we are trying to calculate. The uncertainty in momentum, Δp, is given by h/2, where h is the Planck's constant. Therefore, we can set these two values equal to each other and solve for Δx:

Δx = h/(2Δp)

Now, we need to calculate Δp. We know that p = mv, where m is the mass of the particle and v is its velocity. We also know from the given information that the particles have a diameter of 1.0*10^-6 m. Therefore, we can calculate the mass of each particle using the formula for the volume of a sphere:

m = (4/3)πr^3ρ

where r is the radius of the particle (which is half of its diameter) and ρ is the density of the particle. We can assume that ρ is constant for all the particles, so we can use this formula to calculate the mass of each particle.

Now, we can calculate the uncertainty in momentum, Δp, by multiplying the mass of each particle by the final velocity they acquire during free fall. This will give us the range of possible momenta for the particles.

Finally, we can substitute this value for Δp in the equation for Δx and solve for the diameter of the circle in which most particles will land. This will give us the answer to part A of the problem.

For part B, we need to repeat the same process, but this time using a diameter of 1.7 micrometers for the circle in which most particles will land. We can then calculate the uncertainty in momentum and use it to solve for the distance d at which this diameter will be observed.