= Term symbols for carbon ground state
This example covered for example at <video Term Symbols Example 1 by TMP Chem (2015)>.
<Carbon> has electronic structure 1s2 2s2 2p2.
For term symbols we only care about unfilled layers, because in every filled layer the total z angular momentum is 0, as one electron necessarily cancels out each other:
* <magnetic quantum number> varies from -l to +l, each with z angular momentum $-l \hbar$ to $+l \hbar$ and so each cancels the other out
* <spin quantum number> is either + or minus half, and so each pair of electron cancels the other out
So in this case, we only care about the 2 electrons in 2p2. Let's list out all possible ways in which the 2p2 electrons can be.
There are 3 p orbitals, with three different <magnetic quantum numbers>, each representing a different possible z <quantum angular momentum>.
We are going to distribute 2 electrons with 2 different spins across them. All the possible distributions that don't violate the <Pauli exclusion principle> are:
``
m_l +1 0 -1 m_L m_S
u_ u_ __ 1 1
u_ __ u_ 0 1
__ u_ u_ -1 1
d_ d_ __ 1 -1
d_ __ d_ 0 -1
__ d_ d_ -1 -1
u_ d_ __ 1 0
d_ u_ __ 1 0
u_ __ d_ 0 0
d_ __ u_ 0 0
__ u_ d_ -1 0
__ d_ u_ -1 0
ud __ __ 2 0
__ ud __ 0 0
__ __ ud -2 0
``
where:
* `m_l` is $m_l$, the <magnetic quantum number> of each electron. Remember that this gives a orbital (non-spin) <quantum angular momentum> of $m_l \hbar$ to each such electron
* `m_L` with a capital L is the sum of the $m_l$ of each electron
* `m_S` with a capital S is the sum of the spin angular momentum of each electron
For example, on the first line:
``
m_l +1 0 -1 m_L m_S
u_ u_ __ 1 1
``
we have:
* one electron with $m_l = +1$, and so angular momentum $\hbar$
* one electron with $m_l = +0$, and so angular momentum 0
and so the sum of them has angular momentum $0 + 1 \hbar = 1 \hbar$. So the value of $m_L$ is 1, we just omit the $\hbar$.
TODO now I don't understand the logic behind the next steps... I understand how to mechanically do them, but what do they mean? Can you determine the <term symbol> for individual microstates at all? Or do you have to group them to get the answer? Since there are multiple choices in some steps, it appears that you can't assign a specific term symbol to an individual microstate. And it has something to do with the <Slater determinant>. The previous lecture mentions it: https://www.youtube.com/watch?v=7_8n1TS-8Y0 more precisely https://youtu.be/7_8n1TS-8Y0?t=2268 about carbon.
https://youtu.be/DAgEmLWpYjs?t=2675 mentions that $^{3}D$ is not allowed because it would imply $L = 2, S = 1$, which would be a state `uu __ __` which violates the <Pauli exclusion principle>, and so was not listed on our list of 15 states.
He then goes for $^{1}D$ and mentions:
* S = 1 so $m_S$ can only be 0
* L = 2 (D) so $m_L$ ranges in -2, -1, 0, 1, 2
and so that corresponds to states on our list:
``
ud __ __ 2 0
u_ d_ __ 1 0
u_ __ d_ 0 0
__ u_ d_ -1 0
__ __ ud -2 0
``
Note that for some we had a two choices, so we just pick any one of them and tick them off off from the table, which now looks like:
``
+1 0 -1 m_L m_S
u_ u_ __ 1 1
u_ __ u_ 0 1
__ u_ u_ -1 1
d_ d_ __ 1 -1
d_ __ d_ 0 -1
__ d_ d_ -1 -1
d_ u_ __ 1 0
d_ __ u_ 0 0
__ d_ u_ -1 0
__ ud __ 0 0
``
Then for $^{3}P$ the choices are:
* S = 2 so $m_S$ is either -1, 0 or 1
* L = 1 (P) so $m_L$ ranges in -1, 0, 1
so we have 9 possibilities for both together. We again verify that 9 such states are left matching those criteria, and tick them off, and so on.
For the $m_S$, we have two electrons with spin up. The angular momentum of each electron is $1/2 \hbar$, and so given that we have two, the total is $1 \hbar$, so again we omit $\hbar$ and $m_S$ is 1.
\Video[https://www.youtube.com/watch?v=doC9Z2S7lm8]
{title=Term Symbols Example 1 by TMP Chem (2015)}
{description=Carbon atom.}
Bibliography:
* https://youtu.be/DAgEmLWpYjs?t=1962 from <video Atomic Term Symbols by T. Daniel Crawford (2016)>
Back to article page