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An Electron and Neutron in a Finite Potential Well

  1. Jul 6, 2008 #1
    Given that

    [tex]\frac{2ma^2 |V|}{\hbar^2} = \left(\frac{7\pi}{4}\right)^2[/tex]​

    for an electron in a potential well of depth [tex]|V|[/tex] and width [tex]2a = 10^{-7} \text{cm}[/tex], if a [tex]100\text{-keV}[/tex] neutron is scattered by such a system, calculate the possible decrements in energy that the neutron may suffer.

    We can easily calculate the depth of the potential from the given data... [tex]|V| = 4.60637 \text{eV}[/tex]. Now, if we let [tex]\xi = ka[/tex] while [tex]\nu = \kappa a[/tex], we know that

    [tex]\xi^2 + \nu^2 = \left(\frac{7\pi}{4}\right)^2[/tex]​

    and solving this along with

    [tex]\xi \tan \xi = \nu[/tex]​

    yields a pair of eigenenergies while solving it with

    [tex]-\xi \cot \xi = \nu[/tex]​

    yields another pair of eigenenergies. We solve (using Mathematica, Maple, etc.) and find that the values of [tex]\nu[/tex] can take on [tex]5.3368, 4.8221, 3.8559, 2.0613[/tex], so that means that the eigenenergies take on the values (in milli-electron-volts) [tex]2.36, 1.93, 1.23, 0.35 \text{meV}[/tex]. This means that the neutron may suffer these decrements in energy. [tex]\blacksquare[/tex]

    Are my answers and arguments correct?
  2. jcsd
  3. Apr 22, 2011 #2
    I'm not sure. The mass of the neutron is obviously different from that of the electron, and what you've found is the different energies the electron could take on inside the well. I don't see how this would relate to energy decrements of a neutron scattered by the same system. Maybe someone else on here does?
  4. Apr 22, 2011 #3
    Actually no, I think you're right.

    Assuming the neutron is scattered elastically by the electron in the well, the neutron will lose energy equal to that of the electron. So, this is correct.
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