# Electron spin in an atom

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1. Jul 29, 2015

### Ryan Reed

In the stern-gerlach experiment, silver atoms with a +1/2 would be deflected up, and atoms with a -1/2 spin would be deflected down. With that in mind, would electrons' orbitals within an atom be affected by its spin?

2. Jul 29, 2015

### Staff: Mentor

With an external magnetic field: sure. It is a small effect, however.

3. Jul 29, 2015

### Dr. Courtney

Spin orbit effects are small.

4. Jul 29, 2015

### Ryan Reed

If the electron is the first in a large atom like uranium, would the effect be larger even though it's still small? And does this effect ever get large enough to be measured?

5. Jul 31, 2015

### blue_leaf77

Spin-orbit effect is enhanced as the nuclear charge increases. In heavy atoms such as U the spin-orbit effect is no longer perturbative, and the proper treatment of the problem should start from the relativistic Hamiltonian formula (Dirac equation). The effect due to spin-orbit coupling causes the so-called fine structure effect, where the energy levels are no longer degenerate in $l$, i.e. splitting takes place. For hydrogen like uranium, these splittings might reach keV order of magnitude and yes something that big is hard to miss.

Last edited: Jul 31, 2015
6. Jul 31, 2015

### Ryan Reed

would the spin orbit effect have different results based on which spin was present?

7. Jul 31, 2015

### blue_leaf77

What do you mean by that? The spin of electron is always that $s=1/2$.

8. Jul 31, 2015

### Ryan Reed

An electron's spin can either be either +1/2 or -1/2, that's what the stern-gerlach experiment tested.

9. Jul 31, 2015

### blue_leaf77

It's more accurate to say those numbers as the electron spin projection in a given direction, or the eigenvalues of operator $S_z$.
The presence of spin orbit effect in perturbative regime is expressed as the inclusion of the spin-orbit coupling term proportional to $\mathbf{L}\cdot\mathbf{S}$. This spin-orbit coupling term does not commute with $S_z$, therefore the eigenvalues of $S_z$ are no longer good quantum numbers and we can't really separate the effect of eigenstate with $S_z = \hbar/2$ and that with $S_z=-\hbar/2$. It can be shown that when the spin orbit effect becomes important, one of the good quantum numbers are those corresponding to the total angular momentum operator $\mathbf{J}=\mathbf{L}+\mathbf{S}$.

10. Jul 31, 2015

### Ryan Reed

Ok, but aren't the results of the stern-gerlach experiment due to the positive or negative spin of the outer valence electron? And if so, wouldn't the electron orbitals differ depending on the sign of the spin?

11. Jul 31, 2015

### blue_leaf77

The electron configuration for silver atom is [Kr]4d10 5s1, as you said it has one valence electron in s shell, so the total orbital angular momentum is zero and the total spin angular momentum is 1/2. This makes the total angular momentum $J=S$ and consequently $m_j=m_s$ - the fact that the total orbital angular momentum $L$ is zero makes its effect on the total angular momentum $J$ not apparent and this one is completely governed by $S$. However if a silver atom is free of any perturbation, the energy levels are degenerate in $m_j$. So now answering your question,
it really depends on the system, if the Ag atom is isolated its orbital won't be effected by the sign of the z component of the spin. However, as pointed out in comment #2, if there is external magnetic field, the energy levels are no longer degenerate in $m_j$'s, rather it will split into different energy levels.