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