# 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.
• $2 S + 1$ 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 $2 p$ 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:

$\textcolor{g r e e n}{S} = + \left[\text{1/2"] = color(green)(+"1/2}\right)$

Determine the spin multiplicity, and you should get:

$\textcolor{g r e e n}{2 S + 1} = \textcolor{g r e e n}{2}$

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:

$\setminus m a t h b f \left(J = L + S , L + S - 1 , . . . , | L - S |\right)$

Here, we have:

$J = 1 + \text{1/2", 1 + "1/2" - 1, . . . , |1 - "1/2} |$

but $1 + \text{1/2" - 1 = |1 - "1/2} |$, so we just have two values for $J$.

$\textcolor{g r e e n}{J = \text{1/2", "3/2}}$

OVERALL ATOMIC TERM SYMBOLS

So, we can write out the term symbols as:

""^(2S + 1)L_J

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

$\to \textcolor{b l u e}{\text{^2 P_"1/2", ""^2 P_"3/2}}$

WHAT DOES IT MEAN?

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

1. The number of unpaired electrons is $1$, because $2 S + 1 = 2$, so $S = \text{1/2}$.
2. Because $S = \text{1/2}$, $J - S = L = \text{3/2" - "1/2}$, 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 $2 p$ 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: $\text{^2 P_"1/2}$ and $\text{^2 P_"3/2}$, 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 $\text{1 nm}$ in the wavelength).

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

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