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## Homework Statement

An infinitely deep one-dimensional potential well has a width of 1 nm and contains 10 electrons. The system of electrons has the minimum total energy possible. What is the least energy, in eV, a photon must have in order to excite a ground-state electron in this system to the lowest higher state it can occupy. (Ignore electrostatic interactions between the electrons, but be aware of their fermionic nature.)

## Homework Equations

In an infinite square well, the energy levels are: E_n = n^2*pi^2*hbar^2/(2mL^2) where L is the width of the well, n is the energy level, and m is the mass of the electron.

**3. Attempt at solution**

I think that the electron that will get excited is the 2p^6 one. So the energy transition will be from 2p^6 to 3s^1. So I think I want:

DeltaE = E_3 - E_2 = 5*pi^2*hbar^2/(2mL^2) = 1.88 eV when you plug in the numbers.

What I'm thrown off by is the fact that the problem says a "ground state electron in this system". The ground state electron we would normally think of is the electron in the 1s^1, but it's not going to get excited by a photon because it's in a filled shell. Plus, if we just used the 1s^1 as the electron being excited, then the problem wouldn't have anything to do with the fact that there are 10 electrons in the system. Do others agree that the E_2 to E_3 transition is the one that we are looking for here?

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