Why nuclear spin is ignored in Stern-Gerlach experiment?

Question

I can't sure what isotopes of silver(107 or 109) was used.

But Silver has non-zero nuclear spin since the number of proton and neutron are not even.

Since we use Silver atom not electron in SG experiment,

I think we need to consider spin of electron and silver nucleus simultaneously.

In Sakurai text book, nuclear spin is considered irrelevant to the result of experiment.

If the nuclear spin is 1/2, the total spin of silver atom will be 1.

And we need to observe 3 lines passing through the inhomogeneous B-field.

I don't know why nuclear spin could be ignored.

Answer 1

The separation of the different spin states in the Stern-Gerlach experiment depends on the strength of the magnetic dipole moment. For an electron, this is approximately twice the Bohr magneton:

$$ \mu_e\approx 2\mu_B=2\left(\frac{e\hbar}{2m_e}\right) $$

The nuclear magnetic moment depends on the exact nucleus in question, but it tends to be on the order of the nuclear magneton:

$$ \mu_{\rm nuc}\sim\mu_N=\frac{e\hbar}{2m_p}\ll \mu_e $$

The magnetic fields in a typical Stern-Gerlach experiment done with silver are insufficiently strong to distinguish between atoms of different nuclear spin, so only the electron spin is resolved.