Fermi energy of multiple electrons, infinite potential well

[ricardillo]

Homework Statement

[/B]
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*a²)]*(n_x² + n_y² + n_z²)

The Attempt at a Solution

Tried using E_F = (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)]*(n_x² + n_y² + n_z²) = E₀*(n_x² + n_y² + n_z²).

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 = 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.

 

[blue_leaf77]

Can you give the link to the solution you found?

 

[ricardillo]

Can you give the link to the solution you found?

http://www.slideshare.net/kadubrito5/neamensolution-manual-for-semiconductor-physics-and-devices-3ed

It's on page 25, problem 3.34

 

[blue_leaf77]

Ah ok, sorry I was a bit confused because I forgot that $n_i$ starts from one instead of zero.

EF = (h_bar2/2*m)*(3*pi2*N/V)(2/3

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?

 

[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?

 

[Anik Paul]

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?

I am also facing the same problem.

 

[DrClaude]

I am also facing the same problem.

I don't see how the manual arrived at its solution. The Fermi energy should be at (123).