# Splitting of Lines in Stern Gerlach Experiment

Normally in texts we have read for silver atom for L=0 state, lines split into two parts in Stern Gerlach experiment, discovery of electron spin. What about say 2p state electron for L not equal to zero. Do the lines still split into two parts due to spin or in many parts.

## 1 Answer

The magnetic field $$\boldsymbol{B}$$ couples to the total angular momentum $$\boldsymbol{J} =\boldsymbol{L}+\boldsymbol{S}$$ via a term in the Hamiltonian that is proportional to $$\boldsymbol{J}\cdot \boldsymbol{B}$$.

If there is a non-uniform magnetic field, that roughly points along the $$z$$ axis, you expect a force (see this lecture notes)

$$$$\boldsymbol{F} = \mu_z \frac{\partial B_z}{\partial_z} \hat{\boldsymbol{z}} \propto J_z \frac{\partial B_z}{\partial_z} \hat{\boldsymbol{z}} \hspace{6pt}.$$$$

The total number of possible values of $$J_z$$ (ie the total number of lines you expect to see) is $$2J+1$$, as pointed out in this answer by Sean E. Lake.

The possible values of $$J$$ are constrained by $$|L-S|\leq J \leq |L+S|$$. For a silver atom $$L = 0$$ and $$S=1/2$$, so J is uniquely given by $$J = 1/2$$ and you expect 2 lines.

If the electronic configuration is more complicated, you need to compute all the possible values of $$J$$. If we consider the atom to be in the ground state, $$J$$ can be determined following a set of rules. You can write compactly your configuration using a term symbol $${}^{2S+1}L_J$$, as outlined in the Wikipedia page.

For example, antimony has the following electronic configuration $$[Kr]4d^{10}5s^{2}5p^{3}$$ and its ground state term symbol is $${}^4S_{3/2}$$ (see the NIST website); consequently, $$J = 3/2$$ and we expect $$4$$ lines.