For a cubic box, the energy is:

$\overline{){{\mathbf{E}}}_{{\mathbf{n}}}{\mathbf{=}}\frac{{\mathbf{\u045b}}^{\mathbf{2}}{\mathbf{\pi}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}{\mathbf{L}}^{\mathbf{2}}}{\mathbf{(}}{{{\mathbf{n}}}_{{\mathbf{x}}}}^{{\mathbf{2}}}{\mathbf{+}}{{{\mathbf{n}}}_{{\mathbf{y}}}}^{{\mathbf{2}}}{\mathbf{+}}{{{\mathbf{n}}}_{{\mathbf{z}}}}^{{\mathbf{2}}}{\mathbf{)}}}$, where n_{x} = 1, 2, 3, ......, n_{y} = 1, 2, 3, ...., and n_{z} =1, 2, 3, ....

For a very large n_{rs}:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{{{\mathbf{n}}_{\mathbf{r}\mathbf{s}}}^{\mathbf{2}}{\mathbf{\u045b}}^{\mathbf{2}}{\mathbf{\pi}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}{\mathbf{L}}^{\mathbf{2}}}}$

Problems that require solving the three-dimensional Schrödinger equation can often be reduced to related one-dimensional problems. An example of this would be the particle in a cubical box.

Consider a cubical box with rigid walls (i.e., U(x,y,z)=?outside of the cube) and edges of length L. The general solution for this problem is

where n_{x}, n_{y}, and n_{z} are all positive integers. Note that this solution is just the product of three solutions to the one-dimensional particle in a box. The energy corresponding to the three-dimensional solution is just the sum of the energies for each of the three one-dimensional solutions:

Part A

What is the smallest allowed energy E_{0} for a particle in a cubical box?

Express your answer in terms of ?, m, and L.

Part B

If you are dealing with a very large n_{rs}, you can assume that each state (point with integer coordinates) corresponds roughly to one unit of volume inside of the sphere. So, the number of states is approximately equal to the volume of the octant of the sphere. Use this idea to find the number N of states with energy less than or equal to for a large n_{rs}.

Express your answer in terms of n_{rs}.

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