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## Homework Statement

A system of 20 neutrons are confined in the region

[tex]0<x<3L[/tex]

[tex]0<y<L[/tex]

[tex]0<z<2L[/tex]

With [tex]L=10^{-15}m[/tex]

If each eigenstate can accept 2 neutrons (corresponding to the 2 possible spin states) what is the total kinetic energy of the system?

## Homework Equations

I found the total energy to be

[tex]E = \frac{\pi^{2}\overline{h}^{2}}{2mL^{2}}\left(\frac{(n_{1})^{2}}{9}+(n_{2})^{2}+\frac{(n_{3})^{2}}{4}\right)[/tex]

## The Attempt at a Solution

Now, I do not really understand what the question is asking.

But my current idea is to do trial and error for a combination of ns that gives the same energy and multiply it by 20.

Is this right?

Thanks in advance