# Homework Help: Ground state of 7 electrons in infinite square well

1. Sep 26, 2008

### sabinscabin

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

Seven electrons are trapped in a one dimensional infinite square well of length L. What is the ground state energy of this system as a multiple of h2 / 8mL2?

2. Relevant equations
Energy of a single electron in state n is n2h2 / 8mL2

3. The attempt at a solution

Pauli exclusion principle says all 7 must have different quantum numbers.
starting from n = 1, we have L = 0 and mL = 0, and ms = -1/2 and 1/2, so there are two electrons in n = 1.
For n = 2, we have two electrons for L = 0
for L = 1, we have mL = -1, 0, 1, which means this subshell can hold 6 electrons. The remaining 3 electrons go into this subshell then.

Final tally: 2 electrons for n = 1 and 5 electrons for n = 2.

Total energy as a multiple of the given term then = 2*1^2 + 5*2^2 = 22.

Halliday Resnick says 44 for some reason. Can anybody spot my mistake?

2. Sep 26, 2008

### alphysicist

Hi sabinscabin,

For the 1-D infinite well, there are only two states for each energy level (corresponding to the two spin values).

3. Sep 26, 2008

### sabinscabin

ah thanks. That was incredibly stupid of me. It's in a square well so there's no angular momentum number since there's nothing to orbit.

D'oh!

4. Sep 26, 2008

### alphysicist

No, not stupid. The important thing is to understand why something is wrong. If the test of stupidity is making a mistake, I'm in a lot of trouble.