Is there a general equation that gives the potential energy of any element at each energy level? I know that there is an equation for the potential energy of a hydrogen atom at each energy level but hydrogen is not the only element and from what I understand the potential energy is different for different atoms. How would you calculate the potential energy of oxygen for instance?

Also is there a general equation that gives the classical radius of an atom of each element? How does it affect the classical radius of an atom when the electrons in the outer shell are moved to a higher energy level?


No there isn't. There's an entire branch of physics devoted to calculating those energies: atomic physics. Many people have devoted their entire careers to the task. Even for the next atom, helium, having two electrons (and one nucleus) there is no exact formula.

I'm not familiar with what a classical radius is for an atom.


The hydrogen atom is easy to solve because it's a one body problem (assuming we take the nucleus as fixed). In that case the eigenfunctions of the Hamiltonian are the usual $1s$, $2s$, $2p$, etc.

When we are considering multielectron atoms we make the approximation that we can write the total wavefunction as a product of hydrogenic atomic orbitals, so for boron we might write:

$$ \Psi_B = \psi_{1s}(e_1)\,\psi_{1s}(e_2)\,\psi_{2s}(e_3)\,\psi_{2s}(e_4)\,\psi_{2p}(e_5) $$

where the $\psi_{1s}$ etc are atomic orbitals that resemble the hydrogen atomic orbitals. But there are two key points you need to note:

  1. this is only an approximation because the repulsion between the five electrons causes the atomic orbitals to mix with each other - there are no distinct atomic orbitals

  2. the atomic orbitals are not the same as the hydrogen atomic orbitals. They are not even a simple scaling of the hydrogen atomic arbitals.

To calculate the wavefunction for e.g. boron we would start by using a Hartree-Fock calculation to get a set of atomic orbitals that give the best approximation to $\Psi_B$. Then we would use a configuration interaction calculation to work out how the HF atomic orbitals mix. The end result is a wavefunction $\Psi_B$ that cannot be factorised into separate atomic orbitals.

So there is no simple equation of the type you describe. For all multielectron atoms the calculation of the wavefunction is a complex process (though straightforward on modern computers).

  • $\begingroup$ You don't need to assume nucleus as fixed: the two-body problem in free space is reducible to one-body form in a similar way to classical two-body problem — via reduced mass. $\endgroup$ – Ruslan Jan 19 '17 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.