3
$\begingroup$

I'm trying to understand how to predict ground state term symbol of atoms.

After finding the biggest S, why the biggest L will be $L = |M_L|$, where $M_L = \sum m_l$? I know this rule works to predict the total L, but it doesn't make sense to me that it will give the biggest L, to me it seems like this is the minimum L.

For example, the nitrogen atom has three electrons in 2p.

Following the first rule, the spin $S$ of the ground state will be the maximum possible, which is S = 3/2.

The second rule, according to some books is to maximize $M_L$, in the case of Nitrogen, according to Pauli principle, since all the spins are up, we can only get $M_L = 0$.

Now, all material I read says that L of the ground state will be the maximum one, so L = $|M_L|$, but if we expand the Clebsch-Gordan coefficients of the three states: $|1 \,\,\,\, 0> |1 -1> |1 \,\,\,\, 1>$ we get the possibility of L = 3,2,1 and 0, so why is L = 0 the maximum possible value?

There is an amazing answer to this here, but he doesn't explain why he cuts out the the terms with bigger L.

$\endgroup$
  • 2
    $\begingroup$ Formatting tip: use \langle and \rangle in math mode for bra-ket notation and avoid spamming \,. If you really need space that big, try single \quad or \qquad. $\endgroup$ – andselisk Oct 28 '19 at 18:53
  • 1
    $\begingroup$ Nice tip, thanks. $\endgroup$ – Socrates Oct 28 '19 at 19:25
  • $\begingroup$ I am not sure I understand your question. For the ground state of nitrogen, you are right, you got three electrons on the p orbitals. Then, we maximize then spin, so we get that each orbital is half-filled. The orbitals have magnetic quantum number l=-1,0 and 1 in order. Each of them has one electron, so L=(-1)+0+1=0 and there is no other possibility here. $\endgroup$ – Ezze Oct 29 '19 at 15:08
  • $\begingroup$ P means l = 1 right? So three eletrons with l = 1, means we can get a total momentum of 3, 2, 1, 0. My question is why does it have to be 0. $\endgroup$ – Socrates Oct 29 '19 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.