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Homework Help: Quantum:- SHO potential and its energy

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data

    the energy levels of a particle mass in a symmetric 3d SHO potential are:

    E = (nx + ny + nz + 3/2)*h-bar*[tex]\sqrt{\frac{C}{m}}[/tex]

    n=principal quantum number = nx + ny + nz

    A) If 10 electrons are in the potential whats the lowest possible value for the total energy of all the electrons?

    B) If instead 10 pi- mesons are placed in th same potential what is the lowest possible value of the total energy of the mesons?

    3. The attempt at a solution

    for 10 electrons with + or - 1/2 spin the l quantum number is 2 as if the electrons were in an atom there would be s, p, d orbitals
    This means that n is and integer greater than 2 [lowest n is 3]

    so E = (3+3/2)*h-bar*[tex]\sqrt{\frac{C}{m}}[/tex])
    me = mass of electron

    = (9/2)*h-bar*[tex]\sqrt{\frac{C}{10me}}[/tex])

    i did think that maybe the l quantum number should be 3 but it doesn't need to be in this example it just has 2 as the lowest possible. if all electrons had same spin direction then it would need l=3 due to PEP

    would the answer be the same as above? as l=2 but the mass would be 2730me as pi meson is approx 273 times mass of electron
    = (9/2)*h-bar*[tex]\sqrt{\frac{C}{2730me}}[/tex])
  2. jcsd
  3. Jan 10, 2010 #2


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    Consider that the electron is a Fermion and the pi meson is a Boson. The energy levels they can occupy are different.
  4. Jan 10, 2010 #3


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    The atomic quantum numbers [itex]nlm[/itex] don't apply at all here. That's for an electron in a Coulomb potential. In the 3D SHO, the states are labeled by [itex]n_x[/itex], [itex]n_y[/itex], and [itex]n_z[/itex]. The ground state would be [itex](n_x,n_y,n_z)=(0,0,0)[/itex]. At the next energy level, there are three degenerate states, [itex](1,0,0), (0,1,0), (0,0,1)[/itex], and so on.
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