How to calculate effective magnetic field due to the angular momentum L in an atom like Na(23)? I found an answer that we could imagine the case that the atom was orbiting the electron now and multiply a factor 1/2 to get the initial real case.

so that B = (mu/8*pi)(Ze/m)(L/r^3)

Is this the true case? I use this method and get the answer differ by true case a lot

  • $\begingroup$ Electrons are not orbiting the nucleus, so to calculate the correct magnetic moment you would have to know the actual wave function of that electron: en.wikipedia.org/wiki/Electron_magnetic_moment. $\endgroup$ – CuriousOne Jan 30 '16 at 0:17
  • $\begingroup$ Well, I know how to calculate magnetic moment. What I'm wondering is the effective magnetic field in fine structure. All the book at my hands didn't metion about anything. While the teacher told me something I'm confusing about QQ $\endgroup$ – cindy Feb 6 '16 at 3:39

Your Answer

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

Browse other questions tagged or ask your own question.