What does $p_{1/2}$ and $p_{3/2}$ mean when referring to electron configuration?

I've seen sub shells and I'm assuming it has something to do with spin. I am asking this in regard to the $8p_{1/2}$ and $8p_{3/2}$ shells and the way they get filled at different energy levels?

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    $\begingroup$ See this earlier Q & A, Shape of the P1/2 Orbital $\endgroup$ – ron May 10 '17 at 3:23
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    $\begingroup$ We prefer to not use MathJax in the title field, see here for details. $\endgroup$ – Martin - マーチン May 10 '17 at 5:22
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    $\begingroup$ Essentially, spin-orbit coupling leads to the degeneracy of the 8p orbitals being lifted. ron's question has all the details. $\endgroup$ – orthocresol May 10 '17 at 12:59

Essentially, this derives from Dirac’s relativistic quantum mechanics.

In nonrelativistic quantum mechanics—the Schrödinger equation that you are probably highly familiar with—p orbitals are triply degenerate while d orbitals are 5-way degenerate. However, nonrelativistic quantum mechanics does not include the electronic property known as spin unless it is introduced in a separate step.

Dirac’s relativistic quantum mechanics solves this problem by introducing $c$ as a finite constant. Almost magically, his equations directly result in a quantum number for the spin, $\vec s$. However, the result is that the azimuthal quantum number no longer behaves like a true quantum number (i.e. it no longer describes stationary states) but thankfully the sum $\vec j = \vec l + \vec s$ does. Therefore, instead of considering just one $\vec l$ as a descriptor to decide between s, p and d orbitals, we are stuck with $\left |\vec l - \vec s\right |$ and $\left | \vec l + \vec s\right |$. For $\vec l = 1$ (a p orbital) this gives us the values of $\vec j = \frac12$ and $\vec j = \frac32$. There are two p orbitals of the $\mathrm p_\frac32$ type and one of the $\mathrm p_\frac12$ type. These are no longer degenerate, i.e. $\mathrm p_\frac12$ has a lower energy than the $\mathrm p_\frac32$ ones.

The orbital of the $\mathrm p_\frac12$ type happens to be spherical in nature. For a more thorough background, I recommend Nicolau’s extensive answer which I mostly paraphrased in the above paragraphs and which contains links to further materials.

  • $\begingroup$ Why $p_{3/2}$ has higher energy than $p_{1/2}$? Thanks in advance. $\endgroup$ – ado sar Oct 22 '19 at 18:47

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