# Term symbol - how do we know the number of electrons $e^-$?

Lets say I have a term symbol $^4D_{5/2}$. From this I can simply read the total quantum numbers numbers $L=2$ and $J=5/2$.

Now the superscripted number $4$ is called multiplicity if I am not mistaken which equals:

\begin{align} \text{multiplicity}&=2S+1 \longleftarrow\substack{\text{if L>S}}\\ \text{multiplicity}&=2L+1 \longleftarrow\substack{\text{if L<S}}\\ \end{align}

But how can I know which one to choose as I don't know the $S$ - if I don't know this one I can't say $L<S$ or $L>S$. I could determine the $S$ if I knew how many electrons contribute to the atomic angular momentum $J$ but I don't know how to.

EDIT:

The page from A.Beiser:

• As far as I know, the multiplicity is always $2S+1$, with the term symbol being noted ${}^{2S+1}[L]_J$. Sep 8, 2013 at 20:53
• I copy/pasted the statement in the book. Now tell me if this book is a garbage?
– 71GA
Sep 8, 2013 at 21:42

The superscript in the Russell-Saunders term symbol stands for the spin multiplicity, which is given by $2S+1$. It tells us something about the possible values the z-component of spin can take. In your example, we have to solve the equation $$4=2S+1,$$
which has the solution $S=3/2.$ The possible values of the z-component would be $-3/2$, $-1/2$, $1/2$ and $3/2$ (times $\hbar$). However, it does not tell us the number of electrons in your configuration.
• But in the book A. Beiser, Concepts of modern physics it says that multiplicity is $2S+1$ only in cases when $L>S$ while in cases when $L<S$ the multiplicity is $2L+1$. This is why we can't just state $4=2S+1$ as it could be $4=2L+1$... so can this even be solved?