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Finding the length of a box confining an electron?

  1. Apr 26, 2012 #1
    1. The problem statement, all variables and given/known data

    You want to confine an electron and you want to know for certain that the electron's speed is no
    more than 66 m/s. What is the length of the smallest box in which you can do this?

    A) 2.8 × 10^-6m B) 1.4 × 10^-6m C) 1.1 × 10^-5m D) 5.5 × 10^-6m


    2. Relevant equations

    There are two equations that could be used: E = (1/2m)(hn/2L)^2 or E = (h^2/8mL^2)(n^2), where n = 1, 2, 3, 4...

    3. The attempt at a solution

    I understand that I will be solving for L. However, I don't understand how v = 66 m/s will plug into either of the equations above. Also, I'm not sure how to know what to plug in for n. I know the answer is D.) 5.5 x 10^-6m, but I have no idea how they got that answer. Assistance would be greatly appreciated.
     
  2. jcsd
  3. Apr 26, 2012 #2
    I don't know, the question is weird to me. Without having really worked out the problem myself, I'd just try nonrelativistic kinetic energy relation because it will be quick and easy. If that doesn't work you may have to use a relativistic momentum energy relation.
     
  4. Apr 26, 2012 #3

    collinsmark

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    Gold Member

    I think this is just a simple application of the Heisenberg uncertainty principle.

    There are multiple ways to express the Heisenberg uncertainty principle. One way is in terms of an approximate formula involving uncertainties in terms of "deltas." Another form is a very precise inequality that involves standard deviations, and its use generally requires accurate information about the specific wavefunction shape.

    I'm guessing this problem involves the much easier approximation with the "delta" uncertainties. If you use that one, the answer is one of the listed choices.

    --------------
    Edit: But again, there are different ways to express the Heisenberg uncertainty principle. In its simplest form,

    [tex] \Delta p \Delta x \approx h [/tex]

    But then there is a slightly more approximate inequality,

    [tex] \Delta p \Delta x \gtrsim h [/tex]

    And to make things more confusing, that's sometimes expressed by

    [tex] \Delta p \Delta x \gtrsim \frac{h}{2} [/tex]

    But the most exact version requires you know quite a bit about the wavefunction. You can calculate the variance of the position and variance of the momentum using standard quantum mechanical operators. For example, assuming the expectation value of position and momentum are both 0, then in 1-demension,
    [tex] \sigma_x^2 = \int_{-\infty}^{\infty} \psi^* x^2 \psi \ dx [/tex]
    [tex] \sigma_p^2 = \int_{-\infty}^{\infty} \psi^* \left( -\hbar^2 \frac{\partial^2}{\partial x^2} \right) \psi \ dx [/tex]
    Then in terms of standard deviations, one can show the ultimate in the uncertainty principle:
    [tex] \sigma_x \sigma_p \geq \frac{\hbar}{2} [/tex]
    where [itex] \hbar [/itex] = h/(2π)

    I'm guessing the version you are supposed to use is one of the first three. Check your textbook/coursework for the preferred version in your course.
     
    Last edited: Apr 26, 2012
  5. Apr 26, 2012 #4
    Oh yeah, I think collins is right now that I take a second read. The key wording I missed was "no more," which I guess the problem author thinks can be interpreted as "within" or even (gasp) "uncertain." Sorry for ranting, but if a problem has to resort to language to be hard it's a bad problem.
     
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