27
$\begingroup$

For the Stern-Gerlach experiment done in 1922:

  1. Why were silver atoms used?
  2. Silver atoms contain many electrons in different shells (with different angular momemtum quantum numbers. Why are those not affected by the magnetic field like the $5s$ electron?
Improve this question
$\endgroup$
3
  • $\begingroup$ I've done Stern-Gerlach with sodium myself. I can't say what the difference would be though. $\endgroup$
    – Colin K
    Jul 28, 2012 at 3:37
  • 1
    $\begingroup$ The answer to your first question is persistence. They were methodically working their way through as many metals on the periodic table as they could; silver certainly wasn't the first element they tried. $\endgroup$
    – David H
    Jun 21, 2013 at 11:55
  • $\begingroup$ youtu.be/W-_4zOYTIVU?t=313 also mentions that the Phipps-Taylor experiment (with hydrogen, in 1927) "was more difficult and the effect was less pronounced", but it doesn't say why it was more difficult :-( $\endgroup$
    – Ciro Santilli OurBigBook.com
    Jan 1, 2020 at 10:56

2 Answers 2

Reset to default
22
$\begingroup$

This is a crucial aspect of Stern-Gerlach which is usually omitted from simple descriptions. To see the spin splitting cleanly, you want a massive neutral atom with an unpaired electron. Silver has an unpaired 5s electron, and all others are paired. The 5s electron is in a zero orbital angular momentum state, so as far as the magnetic response is concerned, only the spin changes in different fields.

You use a massive neutral object containing an unpaired S-electron, so that you only split the spin state in the B-field gradient, you don't muck around with orbital magnetic moment inside the atom, or for the particle as a whole. The rest of the silver atom is there to make sure that the only difference in deflection is due to the spin of the outer electron only.

For free electrons, the spin and orbital magnetic moment are related by the Dirac equation, and you can't make the splitting caused by the two different. This is the weird degeneracy in Landau levels, where the spin-splitting is exactly equal to the orbital splitting. This means that the momentum deflections of an electron beam by a magnetic field gradient is always going to be comparable to the deflection due to the different spin states. So you don't split an electron beam cleanly into spin states using magnetic field gradients.

The inner shells of the silver are paired up, so that the electrons are rigid--- you can't mix the electron states in any of the inner orbitals. But the outer electron spin-1/2 makes two degenerate states (ignoring some infinitesimal coupling to the nuclear spin), so the Silver atom ends up being just a massive neutral object with a magnetic moment equal to the magnetic moment due to the spin of a free electron.

Improve this answer
$\endgroup$
3
  • $\begingroup$ But don't the other electrons have orbital angular momentum like these in 2p orbitals ? $\endgroup$
    – Antonios Sarikas
    May 12, 2019 at 23:44
  • 1
    $\begingroup$ Can you elaborate on why the unpaired electron has zero orbital angular momentum? $\endgroup$
    – JosephSanders
    May 16, 2022 at 17:50
  • $\begingroup$ Would you please comment on my related question: Can this experiment give the value of spin? Stern-Gerlach experiment gives the magnetic moment. Then one needs the relation between magnetic moment and spin to find the spin. But this relation is not obvious! $\endgroup$
    – richard
    Feb 9, 2023 at 13:33
8
$\begingroup$

Part 2: as we know today, the "inner electrons" reside in "closed shells". A closed shell has zero total angular momentum and therefore produces onla an extremely weak diamagnetic response - nothing you'd see in the Stern Gerlach experiment.

Part 1: I don't know all the historical reasons, but silver is abundant, chemically fairly stable and easily evaporated, these would make for a number of good practical reasons.

Keep in mind that the spin was not known at the time of the experiment. The intention was to discern between classical (Larmor) theory and the then existing Bohr-Sommerfeld ('old' quantum) theory of electronic motion in atoms. Orbital motion with zero angular momentum also was not known, therefore Stern and Gerlach mistook the splitting in two beams as evidence for orbital quantisation with quantum numbers +1 and -1, respectively. They couldn't have known better at the time.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Wecome to physics SE! Part 1: silver has one valence electron in it's '''5s''' shell. $\endgroup$
    – Stefan Bischof
    Apr 17, 2013 at 12:40
  • $\begingroup$ @hiki: They could have known, because the Sommerfeld quantization always gives odd numbers, so in addition to -1 and 1, you need 0. $\endgroup$
    – Ron Maimon
    Jun 29, 2013 at 22:26

Not the answer you're looking for? Browse other questions tagged or ask your own question.