First of all, I know there are similar related questions but there is something that bothers me.

We can relate the magnetic moment and angular momentum as

$\vec{\mu} = \gamma \vec{l}$ (1)

So in the original experiment, if we don't think about spins (although we consider the orbitals of silver atoms which uses Pauli exclusion principle), silver atoms have zero orbital angular momentum and therefore the only contribution is the spin of the electron in the outermost shell, i.e. $(5s)^1$ electron. Doesn't this also mean that $\mu=0$, therefore, we should expect no deflection at all? But it is said that we should expect a continuous distribution of silver atoms due to their orientations.

In other words, $ F = \mu_z \frac{\partial B_z}{\partial z} = 0$ since $\mu_z = 0$

  • $\begingroup$ Why do you say $\mu$ is zero? $\endgroup$ Mar 5, 2020 at 4:13
  • $\begingroup$ @FellowTraveller Because orbital angular momentum of a silver atom is zero. $\endgroup$
    – Etg
    Mar 5, 2020 at 4:38
  • $\begingroup$ But spin is non zero. This maybe of your interest: plato.stanford.edu/entries/physics-experiment/app5.html $\endgroup$ Mar 5, 2020 at 4:47
  • $\begingroup$ I don't talk about spin. I'm talking about the expected outcome of the experiment where we don't know about spin. Why people expected a continuous distribution of the silver atoms at the first place? Don't we know that angular momentum is zero so that no force will exert on atoms? $\endgroup$
    – Etg
    Mar 5, 2020 at 4:53
  • $\begingroup$ I just read the link you sent. They didn't know that silver atom has zero angular momentum. They were thinking it was L=1. Also they didn't know that L=1 splits into three. So, I guess, this answers my question. Thank you $\endgroup$
    – Etg
    Mar 5, 2020 at 4:58


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