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.)
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?