What is the total angular momentum quantum number?

1 Answer
Jan 18, 2016

#J# is the total angular momentum, which is just a value that collapses #S# and #L# into another variable.


DISCLAIMER: This can be a tough topic, so ask questions if you need to.

ATOMIC TERM SYMBOLS

We should see this in the context of atomic term symbols, which describe:

  • The type of orbital (#s#, #p#, etc)
  • The number of unpaired electrons
  • The possibility for spin-orbit coupling

An atomic term symbol looks like this:

#\mathbf(""^(2S + 1)L_J)#

where:

  • #S# is the total spin angular momentum of all #m_s# for each individual electron in the set of orbitals; it's a fast way of telling you how many unpaired electrons there are.
  • #2S + 1# is called the spin multiplicity, which basically is a more concise way of telling you what #S# tells you, and gives rise to the terminology "singlet state", "doublet state", etc. It's a formal thing.
  • #L# is similar to #l#, which is the orbital angular momentum, i.e. the shape of the orbital.
  • #J# is the total angular momentum, which is just a value that collapses #S# and #L# into another variable.

P1 CONFIGURATION

So, let's take an example. Let's say we had a #2p# orbital with one electron in it. That's known as a #p^1# configuration.

It's the simplest example that isn't too simple:

DETERMINING TOTAL SPIN ANGULAR MOMENTUM

To determine #S#, simply add the #m_s# that you see for each electron. The paired electrons are always going to cancel out.

You should get:

#color(green)(S) = +["1/2"] = color(green)(+"1/2")#

Determine the spin multiplicity, and you should get:

#color(green)(2S + 1) = color(green)(2)#

DETERMINING ORBITAL ANGULAR MOMENTUM

Now, since it's a #p# orbital, #l = 1#, and for the term symbol itself, just as #l# corresponds to the #p# orbital, we write out #L# as #P#.

(Had there been two or more electrons, #L# would not just be as simple as just including #P#. I chose #p^1# for a reason.)

DETERMINING TOTAL ANGULAR MOMENTUM

Finally, #J# is where things get tricky, because there can be multiple #J# values, which ranges from #|L - S|# to #|L + S|#. So, you can have:

#\mathbf(J = L + S, L + S - 1, . . . , |L - S|)#

Here, we have:

#J = 1 + "1/2", 1 + "1/2" - 1, . . . , |1 - "1/2"|#

but #1 + "1/2" - 1 = |1 - "1/2"|#, so we just have two values for #J#.

#color(green)(J = "1/2", "3/2")#

OVERALL ATOMIC TERM SYMBOLS

So, we can write out the term symbols as:

#""^(2S + 1)L_J#

#""^(2S + 1)L_(L pm S)#

#-> color(blue)(""^2 P_"1/2", ""^2 P_"3/2")#

WHAT DOES IT MEAN?

From this, we can work backwards and make the following interpretations:

  1. The number of unpaired electrons is #1#, because #2S + 1 = 2#, so #S = "1/2"#.
  2. Because #S = "1/2"#, #J - S = L = "3/2" - "1/2"#, so #L = 1#, and we are looking at a #p# orbital.
  3. We do NOT know whether there are #1# or #5# electrons total in the three #2p# orbitals because either configuration gives one unpaired electron. But we do know that there are either #1# or #5#, so the possible electron configurations are #p^1# and #p^5#.
  4. We know that in an energy level diagram, we should see two energy states: #""^2 P_"1/2"# and #""^2 P_"3/2"#, which are very close together. Because of an effect called spin-orbit coupling, the two energy levels, which would otherwise be the same, split slightly in a magnetic field (sometimes giving differences of less than #"1 nm"# in the wavelength).

As an example of why this can be important, it tells you that there are two different #"589 nm"# electronic excitations for sodium's singular #3s# electron to an empty #3p# orbital.

http://hyperphysics.phy-astr.gsu.edu/

Both give a yellow emission line upon relaxation, but there are two transitions, not one.