Electron in box considering effects of gravity

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  • #1
mps
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An electron is enclosed in a large rectangular box. Considering gravity, estimate the order of magnitude of the thickness of the layer that is occupied by the electron at the bottom of the box.

I thought about this for a long time and I really have no idea how to proceed. Any thoughts?
 

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  • #2
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Hi mps, welcome to PF. In the future, please do not shirk the problem template: it's there for a reason!

What tools do you have for estimating the spatial extent of a particle? For example, could the uncertainty principle help you out here? Alternatively, could you try and solve for the vertical wave function for the particle under the (approximately linear) potential of gravity?
 
  • #3
mps
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Thanks Steely Dan! I don't know how the uncertainly principle can help here...could you please explain? As for the wave function, does this mean I have to use Schrodinger's equation?
Do I have to use both the uncertainly principle and Schrodinger's or just one or the other?
 
  • #4
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Thanks Steely Dan! I don't know how the uncertainly principle can help here...could you please explain? As for the wave function, does this mean I have to use Schrodinger's equation?
Do I have to use both the uncertainly principle and Schrodinger's or just one or the other?

Well, since you're just being asked for a rough estimate, using the uncertainty principle sounds like the way to go if you can do it; if there's no way to proceed, then one would appeal to the Schrodinger equation (there is a solution to the equation for the gravitational potential, the Airy function). Clearly the electron is more likely to be near the bottom than the top; in particular, all other things being equal, it would prefer to be exactly at the bottom, but the uncertainty principle forbids this. So the "layer" it exists in can be estimated if you can estimate the uncertainty in the momentum. How can this be done? Well, try using the relationship between momentum and energy ([itex]p = \sqrt{2m(E-U)}[/itex]).
 
  • #5
mps
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Thanks so much Steely Dan!
I feel really stupid but how do you know what E and U are?
 
  • #6
mps
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I mean, i know its total energy and potential energy but what are the values?
 
  • #7
vela
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Steely Dan's given you a starting point. You need to make an effort at solving the problem yourself.
 
  • #8
mps
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Sorry but I have thought about it really hard and I really can't come up with anything.
If the electron is happy at the bottom, why would it have any momentum?
If the electron is at the bottom, it has no potential energy... How can we know the kinetic energy K (it's not given)?

p = sqrt (2mK) so to calculate p we need K, but I can't think of a way to get K.

I don't think rest energy helps because that's not mechanical.

Could you please help me some more? thanks!
 
  • #9
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There are two competing effects here: considering the classical potential, the electron would simply sit at the bottom, but considering the uncertainty principle, it cannot be exactly at the bottom, for that would imply zero uncertainty in position (which we know cannot be right). So the electron exists in some layer of width [itex]\Delta x[/itex] where the minimum potential is zero and the maximum potential is [itex]mg\Delta x[/itex].
 
  • #10
mps
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Oh! Well then
[itex]Δp = \sqrt{2m(E-0)} - \sqrt{2m(E-mgΔx)}[/itex]
and we substitute this into ΔpΔx≥h/2∏ to find Δx.

What is E though?
Many thanks! :)
 
  • #11
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Well, what are the energy levels of a well with infinite potential boundaries?
 
  • #12
mps
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Okay En=n22ħ2/(2mL2)

Problems
1. Do we just set n = 1 since when know the particle is near the ground?
2. The equation is derived for when potential energy is zero (when kinetic and potential energy are equal). Is En going to be E in our case? So we get the equation below?*
3. We don't know L. Since the box is "large", L -->∞ so E --> 0?

*
[itex]Δp = \sqrt{2m(∏2ħ2/(2mL2))} - \sqrt{2m(∏2ħ2/(2mL2) - mgΔx)}[/itex]
 

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