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Electron interactions in a Helium ion

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data
    The question was, you have a Helium ion with one electron in its ground state. Then a second electron is added to the n = 10 level. What effect does the second electron have the first electron that is nearest the nucleus? Is the energy increased or decreased?


    2. Relevant equations



    3. The attempt at a solution
    Now I would say the energy would have to be increased, since the first atom is already in its ground state so cant decrease any further. I presume the electrons experience some repulsion from each other, but not sure why this would cause the energy of the first electron to increase.
     
  2. jcsd
  3. Aug 21, 2017 #2

    TeethWhitener

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    Gold Member

    This isn't quite right. You've altered the potential, so the ground states aren't really comparable in this way.

    The total (electronic) Hamiltonian is
    $$H = \sum_{i} \frac{-\hbar^2}{2m_i}\nabla_i^2 - \sum_i \frac{Z e^2}{4 \pi \epsilon_0 \left | \mathbf{R} - \mathbf{r}_i \right | }+\sum_i \sum_{j > i} \frac{e^2}{4 \pi \epsilon_0 \left | \mathbf{r}_i - \mathbf{r}_j \right | }$$
    where the third term on the right side is the electron-electron repulsion. This term is always positive, so electron repulsion always increases the energy of the system (ceteris paribus).

    Edit: I don't think my response above answers your question as posed. Without doing an explicit calculation, I'm pretty certain that the energy of the overall system decreases on addition of the electron. The simple reason is that the ionization potential of helium is positive: it takes energy to remove an electron from helium. The more long-winded explanation is that each of the electrons, at n=1 and n=10, will contribute individually to the Coulomb potential (decreasing energy), but will not have much e-e repulsion, (increasing energy).
     
    Last edited: Aug 21, 2017
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