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Ground state of 7 electrons in infinite square well

  1. Sep 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Seven electrons are trapped in a one dimensional infinite square well of length L. What is the ground state energy of this system as a multiple of h2 / 8mL2?


    2. Relevant equations
    Energy of a single electron in state n is n2h2 / 8mL2


    3. The attempt at a solution

    Pauli exclusion principle says all 7 must have different quantum numbers.
    starting from n = 1, we have L = 0 and mL = 0, and ms = -1/2 and 1/2, so there are two electrons in n = 1.
    For n = 2, we have two electrons for L = 0
    for L = 1, we have mL = -1, 0, 1, which means this subshell can hold 6 electrons. The remaining 3 electrons go into this subshell then.

    Final tally: 2 electrons for n = 1 and 5 electrons for n = 2.

    Total energy as a multiple of the given term then = 2*1^2 + 5*2^2 = 22.

    Halliday Resnick says 44 for some reason. Can anybody spot my mistake?
     
  2. jcsd
  3. Sep 26, 2008 #2

    alphysicist

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    Hi sabinscabin,

    For the 1-D infinite well, there are only two states for each energy level (corresponding to the two spin values).
     
  4. Sep 26, 2008 #3
    ah thanks. That was incredibly stupid of me. It's in a square well so there's no angular momentum number since there's nothing to orbit.

    D'oh!
     
  5. Sep 26, 2008 #4

    alphysicist

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    No, not stupid. The important thing is to understand why something is wrong. If the test of stupidity is making a mistake, I'm in a lot of trouble.
     
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