How Does Quantum Uncertainty Affect Dust Particle Trajectories?

[James98765]

Homework Statement

1.0 micrometer diameter dust particles (m=1.0 * 10^-15 kg) are in a vacuum chamber. The dust particles are released from rest above a 1.0 micrometer diameter hole, fall through the hole, and land on a detector at distance d below.

Part A:
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 = 1.0 m, by how much does the diameter of the circle in which most dust particles land exceed 1.0 micrometers? Is this increase in diameter likely to be detectable?

Part B:
Quantum effects would be noticeable if the detection circle diameter increased by 10% to 1.1 micrometers. At what distance d would the detector need to be placed to observe this increase in the diameter?

Homework Equations

Heisenberg Uncertainty:
dX*dV<=h/2

The Attempt at a Solution

In all honesty, I haven't been able to get started on this problem. I'm not exactly sure how the diameter of the detection circle relates to the vertical distance d.

Should I be using a single slit diffraction analysis to explain the situation?
I considered it but determined that this wasn't the correct model. If single slit diffraction was occurring, wouldn't the dust particles be detected anywhere on the detection screen, not just within a certain diameter.

Does the Heisenberg uncertainty principle explain the larger detection circle diameter?
I believe that dX (uncertainty of position) could be equal to the detection circle diameter. For me to determine dX, I must know dV (uncertainty of velocity). But since I don't know dV, how can I find dX?

Those are basically my two attempts to this problem and both have been unsuccessfull. Am I way off, and if so, can I get a suggesstion? I'm not looking for an entire solution to the problem. I just need a starting point.

Thanks a lot for your help.

 

[phinds]

My fairly uninformed belief is that while the probability of a particle landing at the radius of a HUGE circle is small, it is not zero (and it doesn't matter how big you define HUGE to be, it's just that the bigger it is, the smaller the probability).

 

[Redbelly98]

Homework Equations

Heisenberg Uncertainty:
dX*dV<=h/2

Actually, the Heisenberg uncertainty relates position and momentum, not velocity. So the equation is slightly different than what you wrote.

The Attempt at a Solution

In all honesty, I haven't been able to get started on this problem. I'm not exactly sure how the diameter of the detection circle relates to the vertical distance d.

Should I be using a single slit diffraction analysis to explain the situation?
I considered it but determined that this wasn't the correct model. If single slit diffraction was occurring, wouldn't the dust particles be detected anywhere on the detection screen, not just within a certain diameter.

Does the Heisenberg uncertainty principle explain the larger detection circle diameter?
I believe that dX (uncertainty of position) could be equal to the detection circle diameter. For me to determine dX, I must know dV (uncertainty of velocity). But since I don't know dV, how can I find dX?

dx is the uncertainty in position as the dust particle falls through the 1.0 micrometer hole. Technically, there is zero uncertainty if a sphere falls through a hole of the exact same size, but that would lead to infinite uncertainty in the momentum and this problem would make no sense. So, it seems they want you to use 1.0 micrometers as the uncertainty dx. From that you can use the Heisenberg relation to find the uncertainty in momentum.

 

[James98765]

Thanks for your help. After reading your suggestion I'm relatively certain you're correct.

 

[rwisz]

The Heisenberg uncertainty principle states that it is impossible to simultaneously know both the position and momentum of a particle with absolute certainty. This means that in the case of the dust particles falling through the hole, their position and velocity cannot both be known precisely. This leads to a spread in the landing positions of the particles, resulting in a larger detection circle diameter.

In order to determine the increase in diameter of the detection circle, we can use the Heisenberg uncertainty principle equation, dX*dV<=h/2, where dX is the uncertainty in position and dV is the uncertainty in velocity. Since the particles are released from rest, we can assume that dV is equal to the particles' initial velocity, which is 0. Therefore, we can rewrite the equation as dX*0<=h/2, which simplifies to dX<=h/2. This means that the uncertainty in position is equal to or less than half of the Planck's constant, h.

Now, let's consider the distance d at which the detector is placed. As the distance increases, the uncertainty in position also increases. This means that the detection circle diameter will also increase. In Part A, we are given that d=1.0 m and we need to find the increase in diameter of the detection circle. To do this, we can use the equation for the circumference of a circle, C=2*pi*r, where r is the radius of the circle. Since we are given that the initial diameter of the circle is 1.0 micrometers, we can find the radius by dividing by 2, giving us r=0.5 micrometers. Now, using the equation for the circumference, we can find the increase in diameter by solving for r when C=1.1 micrometers, giving us r=0.55 micrometers. This means that the increase in diameter is 0.05 micrometers, or 5% of the initial diameter.

In Part B, we are given that the increase in diameter is 10% and we need to find the distance d at which this occurs. Using the same equation for the circumference of a circle, C=2*pi*r, we can solve for r when C=1.1 micrometers, giving us r=0.55 micrometers. Now, we can use the equation for the radius of a circle, r=d/2,