# Homework Help: Fermi energy of multiple electrons, infinite potential well

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1. Aug 12, 2015

### ricardillo

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

Five free electrons exist in a three-dimensional infinite potential well with all three widths equal to a 12 angstroms. Determine the Fermi energy level at T 0 K.

2. Relevant equations

E = [(h_bar*pi)2/(2*m*a2)]*(nx2 + ny2 + nz2)

3. The attempt at a solution

Tried using EF = (h_bar2/2*m)*(3*pi2*N/V)(2/3) but no luck; found the solution manual online, but the answer doesn't make sense:

"For a 3D infinite potential well,

E = [(h_bar*pi)2/(2*m*a2)]*(nx2 + ny2 + nz2) = E0*(nx2 + ny2 + nz2).

For 5 electrons, energy state corresponding to nx ny nz = 221 contains both an electron and an empty state, so

EF = E0*(22 + 22 + 12)..." (plug in values and solve from here on)

My question is, why does the 221 state "contain both an electron and an empty state"? It seems like the 5 electrons should fill up only the 111 and 211 levels since 111 has room for two states and 211 has room for six.

2. Aug 13, 2015

### blue_leaf77

Can you give the link to the solution you found?

3. Aug 13, 2015

### ricardillo

4. Aug 13, 2015

### blue_leaf77

Ah ok, sorry I was a bit confused because I forgot that $n_i$ starts from one instead of zero.
This is an approximate formula when the number of electron is very large. For 5 electrons you have to count the possible states one after another starting from the lowest one, which is (111). Neglecting spin, if you have 5 electrons, which level you will end up to if you add the electrons one by one from (111) state?

5. Aug 13, 2015

### ricardillo

If we include degeneracy, then there's one electron in (111), three in (112) and one in (122) with two states left over at that level. However, if I try the same method for thirteen electrons, as in part (b), I'd only get to level (123) rather than the (233) level given by the solution.
Is there something wrong with the way I'm filling up states?