Electron interactions in a Helium ion

In summary: So the question is equivalent to "is the Coulomb potential of a single electron at n=1 greater than the sum of the Coulomb potentials of two electrons at n=1 and n=10?"In summary, the addition of a second electron to the n = 10 level of a Helium ion in its ground state will decrease the energy of the overall system, as the repulsive electron-electron interaction will be outweighed by the decrease in Coulomb potential. This is due to the positive ionization potential of helium, which means that it takes energy to remove an electron from the system.
  • #1
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?


Homework Equations





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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  • #2
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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What are the properties of a Helium ion?

A Helium ion is a positively charged particle that consists of one electron and two protons. It has a mass of 4 atomic mass units and a charge of +2.

How do electrons interact with a Helium ion?

The electrons in a Helium ion interact with the positively charged protons through electrostatic forces. The electrons are attracted to the protons and orbit around them, similar to how planets orbit around the sun.

What is the role of electron interactions in Helium ion stability?

Electron interactions play a crucial role in the stability of a Helium ion. The two electrons in the ion are held in stable orbits around the two protons, creating a balanced electrostatic force and preventing them from colliding with each other.

How do electron interactions affect the energy levels of a Helium ion?

The energy levels of a Helium ion are affected by the interactions between the electrons and protons. These interactions determine the spacing and arrangement of the energy levels, which in turn, affect the ion's chemical and physical properties.

What types of electron interactions are involved in a Helium ion?

The electron interactions in a Helium ion include Coulomb interactions, which are the electrostatic forces between the electrons and protons, and quantum mechanical interactions, which determine the energy levels and properties of the ion.

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