I know that the nucleus of an atom does not contribute to the magnetization of an atom or material, and that the magnetic moments of protons are much weaker than electrons. Why is the magnetic moment of a proton weaker than the electron?

Note: I know that electrons don't actually spin (they're described by wave functions).

  • $\begingroup$ Lazy semi-classical answer: they have the same angular momentum, and the proton is much heavier, so it has a much lower classical angular velocity. $\endgroup$ – Jerry Schirmer Feb 19 '16 at 19:06
  • $\begingroup$ Yes, but I was hoping for some quantum mechanical explanation. Thanks though! $\endgroup$ – PiccolMan Feb 19 '16 at 19:08
  • $\begingroup$ The quantum mechanical answer will come out roughly the same with a lot more work, fwiw. $\endgroup$ – Jerry Schirmer Feb 19 '16 at 19:10
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    $\begingroup$ So in the end, it's all about the mass? $\endgroup$ – PiccolMan Feb 19 '16 at 19:10
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    $\begingroup$ Yes, there are a lot of details in there, but ultimately, you will get a result that depends on the mass and the intrinsic spin. It will be close to the classical picture, but will have some different factors out front. $\endgroup$ – Jerry Schirmer Feb 19 '16 at 19:12

Nuclear magnetic moment is given by $\vec{m_N}=g_N\mu_N\vec{I}$, where $\vec{I}$ is a spin of nucleus, $g_N$ is $g$-factor and $\mu_N=\frac{e\hbar}{2m_p}$ is a nuclear magneton.

Orbital magnetic moment of electron is given by $\vec{m}=-\mu_B\vec{l}$, where $\vec{l}$ is a angular momentum (more precisely - operator of angular momentum) and $\mu_B=\frac{e\hbar}{2m_e}$ is a Bohr magneton.

For proton $\langle\vec{m}_N\rangle\approx 2.793\mu_N $.

Hence ratio $\frac{\langle\vec{m}\rangle}{\langle\vec{m}_N\rangle}\approx \frac{\mu_B}{\mu_N}\approx \frac{2000}{2.793}\approx700$.


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