# Electrons in a box [closed]

According to the Pauli principle, there can be no more than two electrons in a given state. If there are a number N of electrons in this box in the lowest states possible, show that the energy of the "top" electron - the "Fermi energy" $E_F$ is $E_F$ = $(2$$\pi$$\hslash$$N)$$/$$2mA , where A = L^2 and L is the length of the square box the electrons are in. Now what I am thinking of using is (E_{n_x n_y})=(n_x^2+n_y^2)($$\pi^2$$\hslash$$)/$($2mL^2)$. Any hints on proceeding to next steps?

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## closed as off-topic by Nathaniel, Chris White, Dan, akhmeteli, Manishearth♦Jul 11 '13 at 0:21

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Edit. July 8, 2013

Hints:

1. You're on the right track attempting to make integral approximations to count states. In fact, my original first hint above is not as useful as counting states using essentially the procedure you outlined in your first comment.

2. You should be able to show that for large enough energy, namely for large enough $n_x^2+n_y^2$ (or equivalently for large enough $N$, the number of states of the system with values of $n_x$ and $n_y$ satisfying $0<\sqrt{n_x^2+n_y^2}<2mAE/\pi^2\hbar$ is approximately $\pi(n_x^2+n_y^2)/2$. Be careful to include an extra factor of $2$ to account for spin degeneracy.

3. Set the expression for the number of states derived above equal to $N$, and us this to relate $E$ and $N$; this will give the desired expression for the Fermi Energy.

Original Hints:

1. You have written a formula for the eigen-energies as a function of $(n_x, n_y)$; how many states are there for each energy level? In other words, determine the degeneracy of each energy level.

2. Once you've figured out the answer to hint 1, you can now start "filling" up the lowest energy states with electrons making sure not to violate Pauli exclusion.

3. In which energy level is the $N^\mathrm{th}$ electron after having followed hint 2?

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Thanks! For hint 1 I found a graph of $n_x$ by $n_y$ and $n_0$ is drawing a quarter of a circle in first quadrant. It is also clear that $(n_x)^2$ $+ (n_y)^2$ $\leq$ $(n_0)^2).$ $\int 0_(n_0)$ $\int_0_($\pi$/2$)$$n_0 = \pi$$(n_0) ^2/4$. – user26658 Jul 8 '13 at 15:46 sorry about that ^^ I was editing, but I did not know I only had 5 minutes. – user26658 Jul 8 '13 at 15:53 For hint 2, I wrote out the "filled" lowest energy states for n = 7. The energy level for the$7th$electron after hint 2 is$7\$. – user26658 Jul 8 '13 at 16:05