Heisenberg Uncertainty Principle Question

  • #1
CJSGrailKnigh
57
0
a 1.0 micrometer diameter dust particles (mass 1.0x10^-15kg) in a vacuum chamber pass through a hole (diameter assumed to be exactly that of particle) onto a detection plate 1.0m below. By how much does the diameter of the circle increase



Homework Equations



This is clearly a Heisenberg uncertainty question and so,

h/2 = [tex]\Delta[/tex]x[tex]\Delta[/tex]p

or plank's constant over 2 is equal to the uncertainty on x multiplied by the mementum

The Attempt at a Solution



I tried to rearange the uncertainty principle to include the equations of a one dimensional box velocity

v=h/2mL

with the assumtion that the cylinder can be considered a box. However this lead to the uncertainty on x being the separation and that doesn't make any logical sense.

I would appreciate any help on this. Also to give context to the level of depth in theory I have this is a second year Modern Physics course for engineering students.
 

Answers and Replies

  • #2
Andrew Mason
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Perhaps you could give us the exact wording of the question. How fast is the particle moving?

AM
 
  • #3
CJSGrailKnigh
57
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Ya that's the problem I'm having... I can't figure out the velocity of the particle. but here's the exact wording of the problem...

fig 40.48 shows 1.0micrometer diameter dust particles (m = 1.0x10^-15kg) in a vacuum chamber. The particles are released from rest above a 1.0 micrometer diameter hole (there's just enough room for the particles to pass through), and land on a detector at distance d below.

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 particles land exceed 1.0 micrometers.
 
  • #4
Andrew Mason
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Ya that's the problem I'm having... I can't figure out the velocity of the particle. but here's the exact wording of the problem...

fig 40.48 shows 1.0micrometer diameter dust particles (m = 1.0x10^-15kg) in a vacuum chamber. The particles are released from rest above a 1.0 micrometer diameter hole (there's just enough room for the particles to pass through), and land on a detector at distance d below.

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 particles land exceed 1.0 micrometers.
OK. That makes sense now.

Given the mass and velocity of the particle, there would not be much of a quantum effect, unless I am missing something.

I think this can be approached as a particle diffraction problem where the uncertainty as to position is the de Broglie wavelength:

[tex]\lambda = h/p[/tex]

where p is the momentum of the dust particle at the detector. So the diameter of the landing area on the detector is the diameter of the particle [itex]\pm \lambda[/itex]

AM
 
  • #5
CJSGrailKnigh
57
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Hmmmm... that was probably the right way of doing it. A classmate of mine (who is nothing short of brilliant) subbed values obtained using Newtonian results and got what seemed to be a reasonable answer so I used his idea and did the calculations that way. Also the question was made to show how quickly the quantum effects are considered to be negligable (I left that part out). Thanks for the help.
 
  • #6
prakash kumar
16
0
i also agree with Andrew Mason.
he is a good champ.
 

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