An Electron and Neutron in a Finite Potential Well

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SUMMARY

The discussion centers on calculating the energy decrements experienced by a 100-keV neutron when scattered by an electron in a finite potential well with a depth of |V| = 4.60637 eV and width 2a = 10-7 cm. The eigenenergies derived from the equations involving ξ and ν yield values of 2.36, 1.93, 1.23, and 0.35 milli-electron-volts (meV). It is confirmed that the neutron loses energy equivalent to the eigenenergies of the electron due to elastic scattering. The calculations utilize tools such as Mathematica or Maple for solving the eigenvalue equations.

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  • Understanding of quantum mechanics principles, specifically potential wells.
  • Familiarity with eigenvalue problems in quantum systems.
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Domnu
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Problem
Given that

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

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

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

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

and solving this along with

\xi \tan \xi = \nu​

yields a pair of eigenenergies while solving it with

-\xi \cot \xi = \nu​

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

Are my answers and arguments correct?
 
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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?
 
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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