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Ground-state energy of fermions

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the ground-state energy of 24 identical noninteracting fermions in a one-dimensional box of length L? (Because the quantum number associated with spin can have two values, each spatial state can be occupied by two fermions.) (Use h for Planck's constant, m for the mass, and L as necessary.)

    2. Relevant equations

    E=h^2/(8mL^2)[n1+n2+n3+....n24]

    3. The attempt at a solution

    Since the question states that each spatial state can be occupied by two fermions, I thought it would be 48h^2/8mL^2, simplifying to 6h^2/mL^2. However, this is incorrect. Any help would be much appreciated. The fact that two can occupy the same state is throwing me off.
     
  2. jcsd
  3. Oct 11, 2009 #2

    gabbagabbahey

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    I don't understand this equation; I thought the energy levels of a particle in a box were proportional to [itex]n^2[/itex]:

    [tex]E_n=\frac{h^2n^2}{8mL^2}[/itex]

    :wink:

    3. The attempt at a solution
    The fact that two can occupy the same state is throwing me off.[/QUOTE]

    Well, the first two fermions can occupy the [itex]n=1[/itex] state, but the next two will have to go in a higher energy level, [itex]n=2[/itex], and the next two will have to go in the [itex]n=3[/itex] level, and so on...

    So the total ground state energy level will be [itex]E=2E_1+2E_2+\ldots 2E_{12}[/itex], right?
     
  4. Oct 11, 2009 #3
    Thanks. I got it.
     
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