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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.

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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)

\begin{equation} \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}. \end{equation}

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.

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