Stern-Gerlach experiment In the Stern Gerlach experiment, one can determine the value of $j$ (eigenvalue of $J^2$) by counting the number of discrete lines formed on the screen.
For instance, if I count $7$ discrete lines on the screen then I can use the equation $2j+1=7$ to determine the value of $j$. Solving the equation gives me $j=3$. So the value of $m$ (eigenvalue of $J_z$) will be $-3,-2,-1,0,+1,+2,+3$. Clearly, all the lines are formed due to orbital angular momentum because the value of $m$ is an integer.
But what if some of the lines are formed due to spin angular momentum as well?
For example, let the value of $l$ (eigenvalue of $L^2$) be $2$. This provides me $2 \times 2+1=5$ discrete lines on the screen. The particles used in the experiment also have $s$ (eigenvalue of $S^2$) being equal to $\frac{1}{2}$. So the spin of the particles provides me with $2\times\frac{1}{2}+1=2$ discrete lines on the screen. In total, I will again have $7$ discrete lines ($5$ due to $L$ and $2$ due to $S$).
My question is, how will I know in the later case that the value of $j$ is both $2$ and $\frac{1}{2}$ but not $j=3$ (as calculated in the former case)?
 A: The lines are formed by the magnetic field coupling to the total angular momentum $J$, not separately to spin and orbital angular momentum. The way that angular momenta of components add gets quite complicated. The rule is that the $z$-components add ($m_j = m_\ell + m_s$). The rule for the total angular momentum is that $j$ can take take any value that you can form into a triangle with $\ell$ and $s$. In the particular example you give, $j = 5/2$ or $3/2$. If the atom is in a state of $j=3/2$ then you'll see 4 lines. If it's in any other state (mixed or pure $j=5/2$) then you'll see 6 lines.
A: I agree with Sean E. Lake. You need the total angular momentum $J=L+S$ for some atom of interest. Number of blobs is then $2J+1$. For ground state atoms, values of $J$ are tabulated as term symbols:
https://en.wikipedia.org/wiki/Term_symbol
For example, $\mathrm{Ag}$ has $J=1/2$ $\to$ so 2 blobs.
Fe has $J=4$ $\to$ that would give you 9 blobs.
Ca has $J=0$ $\to$ only one central blob, i.e. no SG splitting of the beam.
A: By including spin, one does not add two lines but doubles the number of lines.  Basically every $m$ state is split in two states with $m+1/2$ and $m-1/2$.  Thus in your example with $\ell=2$ and $s=1/2$, one obtains $10$ states.
