Electron interactions in a Helium ion

AI Thread Summary
In a Helium ion with one electron in the ground state and a second electron added at the n = 10 level, the interaction between the two electrons affects the energy of the system. While the first electron experiences some repulsion from the second, the overall energy of the system decreases upon adding the second electron due to the contributions of the Coulomb potential. The ionization potential of helium indicates that energy is required to remove an electron, suggesting that the system stabilizes with the addition of the second electron. The repulsion between electrons does increase energy, but its effect is outweighed by the stabilization from the Coulomb interactions. Thus, the addition of the second electron ultimately leads to a decrease in the overall energy of the Helium ion system.
Cassidyzialle
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Homework Statement


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?


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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 can't 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.
 
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Cassidyzialle said:
Now I would say the energy would have to be increased, since the first atom is already in its ground state so can't decrease any further.
This isn't quite right. You've altered the potential, so the ground states aren't really comparable in this way.

Cassidyzialle said:
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.
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).
 
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