# 8 Electrons in a 3-D Infinite Square Well w/ Spin

1. Aug 16, 2013

### klawlor419

1. The problem statement, all variables and given/known data

A cubical box whose sides are length L contains eight electrons. As a multiple of $$\frac{h^2}{2mL^2}$$ what is the energy of the ground state of the eight electrons?

Assume the electrons do not interact with each other but do not neglect spin.

2. Relevant equations

1.) $$E_n=\frac{h^2}{2mL^2}(n_{x}^2+n_{y}^2+n_{z}^2)$$
2.) The Pauli Exclusion Principle

3. The attempt at a solution

My logic is that there are only two electrons allowed in each energy level by the Pauli Exclusion principle. Therefore, two electrons will occupy each of the single electron quantum states. This sounds like a terrible assumption but regardless I just decided to add up 2 times each of the first four energy levels of the single particle quantum square well.

$$K=\frac{h^2}{2mL^2}$$
$$E_{0}=\text{Energy of the ground state of the 8 electrons}$$
$$E_0=2\cdot 3K+2\cdot 6K+2\cdot 9K+2\cdot 12K=60K$$

I got the factors of 3,6,9 and 12 from: $$n_{x}^2+n_{y}^2+n_{z}^2$$

The 1,1,1 produces 3
The 2,1,1 produces 6
The 2,2,1 produces 9
The 2,2,2 produces 12

Did I do this correctly? Thanks ahead of time.

2. Aug 16, 2013

### TSny

What about 1,2,1, etc?

EDIT: Are you sure the numerical factor of 2 in the denominator is correct in your expression $$E_n=\frac{h^2}{2mL^2}(n_{x}^2+n_{y}^2+n_{z}^2) \; ?$$

Last edited: Aug 16, 2013
3. Aug 16, 2013

### klawlor419

I see now, I knew I was slightly missing something! Thanks for catching the factor also. I had it on paper as an 8 ;)

So the first four states that get filled with 2 electrons each, are the:
1,1,1
2,1,1
1,2,1
1,1,2

Though the last 3 are states that have the same energy, they are distinct states that need to be filled before the:
2,2,1
2,1,2
1,2,2 states get filled!

Thanks!

4. Aug 16, 2013

Good work!
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