On my last test, I had trouble with a couple questions because I wasn’t totally clueful about how angular momentum works in atoms. I should fix this.
Orbital angular momentum should come naturally: Our standard way of understanding electron states in atoms is with orbitals, which are chosen to have well-defined angular momentum quantum number l and magnetic quantum number m. (Essentially, l is the magnitude of the angular momentum vector and m is the z-component of the angular momentum. Of course this is kind of a lie but whatever.)
But every electron has not only orbital angular momentum from its motion around the nucleus, but also spin angular momentum, which is intrinsic. This spin always has ls=1/2, so along any given axis ms=±1/2.
The question is: what happens when we add l+s? That is, what is the total angular momentum of the electron? We’ll call this j. By looking at vector magnitudes, we see that the angular momentum quantum number j must lie between \(l+\frac{1}{2}\) and \(|l-\frac{1}{2}|\) (note the careful absolute value…). Since j has to be an integer or half-integer, we see that we must have j=l±1/2, unless l=0 in which case j=1/2. For each of these j values, we have the standard set of mj values from +j to -j.
What if we have an atom with a bunch of electrons in it? What does its angular momentum look like? Well, if a subshell (fixed n & l) is full, we know that it contributes neither orbital nor spin angular momentum — everything cancels out. So we need only look at partially-filled subshells. We fill such subshells according to Hund’s rules. Rules 1 & 2 tell you how to place your spins, rule 3 tells you how to get a total angular momentum J from this.
Ex: a P subshell
| ml | L | S | J | ||||
|---|---|---|---|---|---|---|---|
| +1 | 0 | -1 | |||||
| 2P1/2 | ↑ | 1 | 1/2 | 1/2 | (less than half filled; lowest J value is lower-energy) | ||
| 3P0 | ↑ | ↑ | 1 | 1 | 0 | (“) | |
| 4S3/2 | ↑ | ↑ | ↑ | 0 | 3/2 | 3/2 | (half filled; L=0, only one possibility) |
| 3P2 | ↑↓ | ↑ | ↑ | 1 | 1 | 2 | (more than half filled; highest J value is lower-energy) |
| 2P3/2 | ↑↓ | ↑↓ | ↑ | 1 | 1/2 | 3/2 | (“) |
| 1S0 | ↑↓ | ↑↓ | ↑↓ | 0 | 0 | 0 | (“) |
We denote electron configurations with term symbols. They look like 2S+1LJ. L is given in “spectroscopic notation” (S/P/D/F/etc.). The “2S+1″ part is called spin multiplicity. A singlet state is one with total spin 0, a doublet state is one with total spin 1/2, a triplet state is one with total spin 1, etc. We’re counting the number of possible mS values here, which if l=0 is the number of split energy levels in the presence of a magnetic field.
Example: The base state of neutral helium is 1s2, with spins like ↑↓ in the 1s subshell. This has zero total spin and zero total orbital spin, so it’s 1S0, a singlet. The first excited state of helium is 1s12s1, with two up spins. This has total spin 1 and zero total orbital spin, so it’s 3S1, a triplet. The next excited state is 1s↑2s↓ with term symbol 1S0, a singlet. Hund’s rules are what tell us this is a higher-energy excited state.