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Energy required to remove all three electrons from Li

  1. Oct 8, 2005 #1
    How do you calculate the energy required to remove all three electrons from a Li atom? Here is what I have come up with so far.

    I know Bohr's formula E_n = -13.6 Z^2/n^2. There is one electron in the outer orbital (n = 2), and two in the inner orbital (n = 1).
    E_2 = -13.6*9/4
    E_1 = -13.6*9/1

    So 2*E_1+E_2 = 20.25*-13.6 = 275.4

    Is this correct?

  2. jcsd
  3. Oct 8, 2005 #2

    Physics Monkey

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    Your model treats the electrons as non-interacting, but the electrons do have a Coulomb interaction between them. The experimentally measured energy required to liberate all three electrons is 202.5 eV. The calculation you have done represents, if you like, a kind of zero order estimate, but clearly this doesn't cut it. It is the business of theoretical atomic physics to try to improve on the non-interacting electron estimate.

    So unfortunately there is no simple answer to your question. The Schrodinger equation with even two electrons (He) is already impossible to solve exactly. A good place to learn more about the techniques of atomic physics is the book "Physics of Atoms and Molecules" by Bransden and Joachain.
    Last edited: Oct 8, 2005
  4. Oct 8, 2005 #3
    Thanks for your response. Is there a simple first order approximation that does better than treating the electrons as non-interacting?
  5. Oct 8, 2005 #4


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    Be careful here. For the 1st electron removed, the Zeff would be +1, for the second electron removed Zeff would be +2, but for the third electron removed Z = +3, so one cannot simply double E_1. For each electron, the Bohr energy will be different because the Zeff changes as each electron is removed.
  6. Oct 8, 2005 #5

    Physics Monkey

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    The simplest thing you can try is first order perturbation theory, but that doesn't work very well. About the easiest method with any reliability is a variational approach. As Astronuc indicated, the effect of the other electrons can be seen as shielding or screening the outermost electron from the full charge of the nucleus. If you use a variational parameter corresponding to Z you can sometimes get decent in-the-ballpark results. This often works better than usual when calculating the first ionization energy of the alkalis since they have a single electron outside a closed shell. The trial wavefunctions get rather complicated however as you add lots of electrons, three would already be pretty bad. The procedure here would be to estimate the energy required to remove the first and second electrons using two variational calculations, and then the final ionization energy is given by the Bohr formula. I don't know how well it will work though, I've never tried to compute atomic structure for Li before. If memory serves, the first ionization energy is something like 5 eV.

    More sophisticated and systematic approaches have been developed, and you can read about them in any atomic physics book.
  7. Oct 9, 2005 #6


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    If this is a school homework problem that you are required to solve using the Bohr model, your answer is correct.

    In reality, it is only an order of magnitude estimate, and will be an overestimate (because, as has been explained above, it doesn't count shielding/screening which makes the effective nuclear charge smaller; see Slater's work on effective nuclear charge). The real value of the 3rd ionization energy of Li is less than half your number - it's about 120 eV.

    See webelements for the correct value (remember to convert; the conversion is roughly a factor of 100).

    Slater's rules for calculating effective nuclear charge
  8. Oct 9, 2005 #7
    1st ionization state: 5.3917 V
    2nd ionization state: 76.638 V
    3rd ionization state: 122.451 V

    hope this helps.
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