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In the Stern-Gerlach experiment, they used a beam of silver atoms travelling in z direction (which has valence electron in the 5s shell), so the electrons carry zero angular momentum. Now since they have zero angular momentum, their magnetic moment must be zero. But when experiment was done by Stern and Gerlach, they expected a gaussian distribution because the magnetic moments were randomly oriented.

But my question is, even if the magnetic moments are randomly oriented, they're still zero, so if hypothetically there were no spin quantization, then we must get a single intense point on the screen because all the electrons must travel in same direction, undeflected by the spatially varying magnetic filed?

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Stern and Gerlach didn't know that electrons carry spin, so they did in fact think that their intrinsic magnetic moment is 0. But through experiments they still knew that the magnetic moment of a silver atom was non-zero. They attributed this to an orbital angular momentum with quantum number $l=1$, though, so they expected three dots on the screen (one each for the quantum numbers $m_l=-1,0,+1$), not two. So the reason for the classically expected Gaussian distribution at the time was a supposed orbital angular momentum.

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  • $\begingroup$ They thought that valence electron in silver atom will not have $l=0$ state and will be in $ll=1$ state? @Vercassivelaunos $\endgroup$ – Shine kk Jul 13 at 6:24
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The electrons do not have zero angular momentum; they have zero orbital angular momentum, meaning that the total angular momentum of the electrons (and to a good approximation, the silver atoms) is given by the intrinsic angular momentum (spin) of the electrons, which was assumed to be uniformly distributed across all directions. Thus, the magnetic moment is non-zero and proportional to the electron spin, causing the atoms to spread in the magnetic field.

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  • $\begingroup$ What other experiment or calculations they did before the Stern-Gerlach experiment, which ket them know that there is some other non zero angular momentum in addition to orbital angular momentum? @Sandejo $\endgroup$ – Shine kk Jul 13 at 6:12
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    $\begingroup$ @Shinekk: None, because they didn't know. They thought that the orbital angular momentum was non-zero. $\endgroup$ – Vercassivelaunos Jul 13 at 6:19
  • $\begingroup$ You mean to say that, they thought $l=o$ state will have non zero angular momentum, against the hydrogen atom problem which suggested that $l=0$ state will have zero angular momentum? @Vercassivelaunos $\endgroup$ – Shine kk Jul 13 at 6:21
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    $\begingroup$ @Shinekk: No, they thought the ground state was the $l=1$ state. $\endgroup$ – Vercassivelaunos Jul 13 at 6:25
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    $\begingroup$ @Shinekk: Exactly, and they did expect that. So only seeing two dots was a surprise to them. $\endgroup$ – Vercassivelaunos Jul 13 at 6:30

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