Finding the Hamiltonian for the Be Atom without Nuclear Terms

In summary, the conversation discusses using the QM Hamiltonian for the beryllium atom and difficulties in finding it using the Born-Oppenheimer approximation. The person seeking help has questions about the nucleus being treated as a point particle charge and the inclusion of electron spin. The expert suggests adding all possible terms, including kinetic energy for the electrons and nucleus, Coulomb interaction, and spin terms. The person seeking help thanks the expert and mentions being able to leave out the nucleus in the approximation.
  • #1
rovert
7
0
I have a problem that uses the QM Hamiltonian for the berylium atom, but I am having trouble finding this Hamiltonian using the Born-Oppenheimer approximation (leaving out the nuclear-nucler and nucler-electron terms).
Any know how to get this?
 
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  • #2
It's not difficult. How many electrons does this atom have ? Can you treat the nucleus as a point particle charge ? Are you requires to use the spins of the electrons?

Daniel.

P.S. You can't leave out the nuclear-electron interaction terms.
 
  • #3
Be has 4 electrons. Yes, they are assumed point charges. Electron spin should be included. Does it start with H=(-ћ2/2m)∑…? That is where I thought I should start, but am stuck after that. I have never seen a Hamiltonian developed for a many-electron atom.
 
  • #4
You just have to add all possible terms: KE for the 4 electrons, KE for the nucleus, then Coulomb interaction for the 4 electrons among each other , Coulomb interaction for the 4 electrons with the nucleus and finally spin terms: spin-orbit interaction for the electrons and spin-spin interactions for the electrons, that is, of course, if you neglect nuclear spin.

Daniel.
 
  • #5
Thanks for your help. I think I've got it now. Since I can make the Born-Oppenheimer approximation, I can leave KE for the nucleus out (basically ignore the nucleus all together) which I know is a HUGE assumption to make, but that is what the problem instructs.
 

Related to Finding the Hamiltonian for the Be Atom without Nuclear Terms

1. What is the Hamiltonian for the Be atom?

The Hamiltonian for the Be atom is a mathematical operator that describes the total energy of the atom. It includes terms for the kinetic energy of the electrons and the nuclear potential energy.

2. How is the Hamiltonian for the Be atom derived?

The Hamiltonian for the Be atom is derived from the Schrödinger equation, which is a fundamental equation in quantum mechanics that describes the behavior of particles at the atomic scale. It takes into account the interactions between the electrons and the nucleus of the atom.

3. What does the Hamiltonian for the Be atom tell us about the atom?

The Hamiltonian for the Be atom provides information about the energy levels of the atom, as well as the probability of finding an electron in a particular state. It also allows us to calculate other properties of the atom, such as its magnetic moment and chemical reactivity.

4. How does the Hamiltonian for the Be atom differ from that of other atoms?

The Hamiltonian for the Be atom is unique to this specific atom and differs from that of other atoms due to the number of protons, neutrons, and electrons in the atom. This results in different energy levels and electron configurations for each atom, leading to a different Hamiltonian.

5. Can the Hamiltonian for the Be atom be solved exactly?

No, the Hamiltonian for the Be atom cannot be solved exactly due to the complexity of the atom's structure and interactions. However, numerical approximations and advanced computational methods can be used to calculate and approximate the Hamiltonian for the Be atom and other atoms.

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