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Homework Help: Approximate the minimum kinetic energy of an electron

  1. Oct 6, 2005 #1
    Q: Approximate the minimum kinetic energy of an electron confined to a region the size of an atom (0.10 nm)

    How would one go about solving this? What equations/principles are involved and how do we use them?

    Seems like a simple question but i have no idea where to start? :confused:

    Thanks in advance!
  2. jcsd
  3. Oct 6, 2005 #2

    Andrew Mason

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    Use the Heisenberg Uncertainty principle to determine the momentum of the electron. From that momentum, determine its kinetic energy.

  4. Oct 6, 2005 #3
    How does that work? I've asked this before. The HUP will give you the uncertainty in momentum in this case - i.e. the difference between the minimum and maximum momenta. I don't see how it gives you the actual minimum momentum.

    If the electron has no probability of being found outside this confined region (i.e. an infinite well), then you can model it as having a potential energy of 0 inside and infinity outside. The minimum kinetic energy is then the total energy of the electron for the n = 1 state. However, that depends on an assumption not given by the OP.
  5. Oct 6, 2005 #4

    Andrew Mason

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    In order to contain a particle with mass [itex]m_e[/itex] in a region of space of volume [itex](\Delta x)^3[/itex], the uncertainty principle applies:

    [tex]\Delta x \Delta p > \frac{\hbar}{2}[/tex]

    This means that the range of speed is given by:

    [tex]\Delta v > \frac{\hbar}{2m_e\Delta x}[/tex]

    As [itex]\Delta x[/itex] becomes very small, the uncertainty in its speed becomes very large. If the actual speed was known to be less than [itex]\Delta v[/itex] while we knew that it was still confined to that space, the Uncertainty Principle would be violated. Therefore, the actual speed must be greater than [itex]\Delta v[/itex]. This means that there is a minimum energy that the electron must have:

    [tex]E = \frac{m_ev^2}{2} > \frac{\hbar^2}{8m_e\Delta x^2}[/tex]

    Last edited: Oct 6, 2005
  6. Oct 6, 2005 #5
    Ahhhhh, of course! So obvious now it's been explained to me. So the HUP, as well as limiting the uncertainties, tells us that the maximum position is also given by [itex]\Delta x[/itex] and the minimum momentum is given by [itex]\Delta p[/itex].

    This is the second utterly simple and highly useful thing I've found out on PF in the last 16 hours that I haven't been taught. I should jack in uni and stay on PF 24/7.
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