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Homework Help: Energy levels of Lithium. Quantum.

  1. Dec 7, 2009 #1
    I am having an issue calculating the first few energy levels of Lithium. To add, electron-electron repulsions are ignored.

    Since electron-electron repulsion is being ignored, I am starting with the following,

    For Hydrogen,


    Where the ground state, n=1,


    As mentioned, since we are ignoring electron-electron repulsion, the only difference will the amount of charge.

    So substituting,


    So for Lithium,


    But we know,




    With this we can conclude that,



    [tex]E_t=9\left(\frac{E_1}{n^2}+\frac{E_1}{n'\ ^2}+\frac{E_1}{n''\ ^2}\right)[/tex]

    Now here is where my question comes in. I need to find first five energy levels of Lithium. I am a little confused on to which values of n, n', and n'' to choose.

    A friend of mine said that first five states would be, 1 1 2, 1 1 3, 1 1 4, 1 1 5, and 1 1 6.

    I on the other hand think it should be, 1 1 2, 1 2 2, 1 1 3, 2 2 2, and 1 2 3.

    Can someone shed some light? I always thought to go by the sums of the squares of each n combination. So my order follows chronologically from lowest to the next possible energy level. His on the other hand, seem to skip one's that I think should be present.

    All help is appreciated!
    Last edited: Dec 8, 2009
  2. jcsd
  3. Dec 8, 2009 #2
    E_t = 9\left(\frac{E_1}{n^2}+\frac{E_1}{{n^{\prime}}^2}+\frac{E_1}{{n^{\prime \prime}}^2}\right)

    You have a nice formula for the energy. Why don't you evaluate it for different n, n' and n'' values and compare them with one another?
  4. Dec 8, 2009 #3
    That is what I'm doing.

    The states that are listed above, "1 1 2, 1 2 2, 1 1 3, 2 2 2, and 1 2 3", I plugged into that equation. The first number being n, second, n' and third being n''.

    For example, 1 1 2, the ground state would be,

    [tex]E_t = 9\left(\frac{E_1}{1^2}+\frac{E_1}{{1}^2}+\frac{E_1}{2^2}\right)= 9\left(\frac{E_1}{1}+\frac{E_1}{1}+\frac{E_1}{4}\right)= \frac{81E_1}{4}= \frac{81(-13.6\ eV)}{4}= -275.4\ eV[/tex]

    I understand how to use the formula.

    Well let me ask this, do I need to use combinations of n, n', and n'' such that the energy levels increased minimally from one another?

    What I think should be true is that E2 cannot be say 300eV if a combination of n's allows an energy of 295eV. So 295eV should be E2, not 300eV.

    Get where I'm coming from? I just need to know if the last statements are true.
  5. Dec 8, 2009 #4
    Bummer. I made a careless error. What a waste of time typing.

    I, like a knuckle head did not actually try the states I suggested. I was looking at it as,

    [tex]E = \left({n^2}+{{n^{\prime}}^2}+ {n^{\prime \prime}}^2}\right)[/tex]

    rather than,

    [tex]E = \left(\frac{1}{n^2}+\frac{1}{{n^{\prime}}^2}+ \frac{1}{{n^{\prime \prime}}^2}\right)[/tex]

    I was careless. So 1 1 2, 1 1 3, 1 1 4, 1 1 5, and 1 1 6 are the correct ones.

    Thanks for the help though.
  6. Dec 8, 2009 #5
    I'm sort of confused. If you thought it was [tex]E = \left({n^2}+{{n^{\prime}}^2}+ {n^{\prime \prime}}^2}\right)[/tex], then wouldn't the lowest states be n=infinity, as then you'd get a huge negative number.
  7. Dec 8, 2009 #6
    I hear ya on that part.

    I was thinking the same thing, and am going to ask the good ol' professor tomorrow.
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