What is the magnitude of the B field produced by the electron's orbital motion in H1, according to classical model? I found all sorts of values, I myself calculated 1/8 T, using the standard formula, but I suppose it is not exact.

What is QM response on the issue? Since in that theory the electron has no defined orbit I imagine it cannot produce a magnetic field. Can you explain what is the QM explanation of the spin-orbit interaction and what is the presumed / measured value of the B field in the ground state, is it zero?


The ground state of hydrogen is an S state and has an orbital angular momentum of zero. So the only magnetic moments present will be those due to the spins of the electron and the proton. These spins can either align to give a total spin of one or anti-align to give a total spin of zero. The magnetic moment is related to the total spin by:

$$ \mu = g\mu_B\sqrt{s(s+1) } $$

So when the electron and proton spins are aligned the magnetic moment is $2\sqrt{2}\mu_B$ while when the spins are anti-aligned it is zero.

Incidentally, the transition between the two states is responsible for the hydrogen line.

| cite | improve this answer | |
  • $\begingroup$ @ally: I've never done the classical calculation and I'm afraid I have no interest in doing it since it is of no physical relevance. $\endgroup$ – John Rennie Jun 9 '16 at 10:54
  • 1
    $\begingroup$ I don't think the Bohr magnetron can be applied to a proton in the way shown in this answer. The much smaller nuclear magnetron should be used for the proton. The proton and electron should be considered as two separate dipoles as: pha.jhu.edu/~rt19/hydro/node9.html $\endgroup$ – DavePhD Oct 26 '16 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy