What happens to an atom's angular momentum when it absorbs an electron?

[LCSphysicist]

Homework Statement

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Relevant Equations

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I was thinking a little about how the absorption of angular momentum occurs from the point of view of QM. For example, suppose we have an atom A and an electron $e^-$.

The electron $e^-$ is ejected from a source radially in direction of the center of the atom. Suppose that the atom has net angular momentum $ = 0 $ and it absorbs the electron, my question is what will happen now.

I mean, the electron has angular momentum $\hbar /2$, but since it was emitted from a source in a randomly way, **it is equally probable that the spin of the electron (if we measure it while it is on the path to the atom) be collapsed in any direction.**

So, my interpretation is that the atom, after absorbing the electron, will maintain its angular momentum = 0, since in an average way the electron has angular momentum = 0.

Now, suppose we create a source that emits electron in such way that the direction of the spin is the z axis. The electron now is in the state $|\psi \rangle = |+ \rangle/\sqrt{2} + |- \rangle/\sqrt{2}$

Supposing that the atom is enclosed by a box with an open hole in its surface that allows the electron to pass, *but that does not allow us to see inside it. Is it right to say that the atom now is in the state $|\psi \rangle = |+ \rangle/\sqrt{2} + |- \rangle/\sqrt{2}$? In other words, it is equally probably the atom angular momentum $\hbar/2$ in the + or - direction?*

I would like to know if my statement in bold is right, and if it is right to interpret the italic way.

 

[kuruman]

Consider a sodium ion (Na⁺) that has zero spin and orbital angular momentum, $S=0$ and $L=0$. Its total angular momentum is $J=L+S=0.$ Say it captures an electron. The atom now has $S=\frac{1}{2}$ and $L=0.$ Its total angular momentum is not zero, it is $J=L+S=\frac{1}{2}.$ If you have a collection of such atoms and you pass them through a Stern-Gerlach machine, ideally half of them will come out "spin up" relative to the machine and the rest "spin down".