I am trying to calculate the orbital magnetic moment, $\bar{\mu}$, for Sodium, which has an electron configuration of $1s^2 2s^2 2p^6 3s^1$. The full shells do not contribute to $\bar{L}$ and $\bar{S}$ so only the outer shell electron will contribute to them.

The $3s$ shell corresponds to $l=0$. And $\bar{L}=\hbar\sqrt{l(l+1)}$ which gives $\bar{L}=0$.

Is this possible? Doesn't that mean that the electron isn't "oribiting" the nucleus? I am thinking that $\bar{L}^2$ tells you if the electron is "orbiting" the nuclues, but then what is the significance of $\bar{L}$?

Extra question:

The z-component of the orbital magnetic moment is given by $\mu_z = -m_l\mu_b$. Since $m_l$ varies as $-l\le m_l\le l$ does that mean that there are more than one values for the z-component? Strange, no?

  • $\begingroup$ Electrons don't orbit the atom like planets orbiting a star. The electrons in atoms exist as a delocalised probability distribution. This distribution can have a zero or non-zero angular momentum, but the latter can't be simply interpreted as the electrons circling the nucleus. $\endgroup$ – John Rennie Feb 12 '14 at 15:30
  • $\begingroup$ @JohnRennie You're right, I was being lazy. I added speech marks to make it a little bit more formal :) $\endgroup$ – turnip Feb 12 '14 at 15:36
  • $\begingroup$ My point was that $L$ is just the angular momentum of the atomic orbital, and it can indeed be zero so the total orbital angular momentum of a sodium atom is zero. Why is it strange that $L_z$ can vary from $-L$ to $L$? $\endgroup$ – John Rennie Feb 12 '14 at 15:41
  • $\begingroup$ Well I was more concerned with $\mu_z$. Since $m_l$ varies between $-l$ and $l$, I see why there can be a sign change but in cases where $l=2$ you can get four values (two for the positive "side" and two for the negative). What does that mean in physical terms for the electron? $\endgroup$ – turnip Feb 12 '14 at 15:49
  • $\begingroup$ For $l = 2$ there are of course five values; you missed $l_z = 0$. Anyhow, the $l_z$ value just gives the angle the magnetic dipole makes with the $z$ axis i.e. it's the $z$ component of $l$. Classically this angle is continuous but in QM it's quantised. $\endgroup$ – John Rennie Feb 12 '14 at 15:58

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