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## Homework Statement

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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.

## Homework Equations

E = [(h_bar*pi)

^{2}/(2*m*a

^{2})]*(n

_{x}

^{2}+ n

_{y}

^{2}+ n

_{z}

^{2})

## The Attempt at a Solution

Tried using E

_{F}= (h_bar

^{2}/2*m)*(3*pi

^{2}*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*a

^{2})]*(n

_{x}

^{2}+ n

_{y}

^{2}+ n

_{z}

^{2}) = E

_{0}*(n

_{x}

^{2}+ n

_{y}

^{2}+ n

_{z}

^{2}).

For 5 electrons, energy state corresponding to n

_{x}n

_{y}n

_{z}= 221 contains both an electron and an empty state, so

E

_{F}= E

_{0}*(2

^{2}+ 2

^{2}+ 1

^{2})..." (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.