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In Sakurai's Modern Quantum Mechanics, when he trying to derive the magnetic moment of the silver atom, he simply the whole atom have the same moment as the 47th electron's spin magnetic moment and stated that it's proportional to the electron spin S and the factor is $e/m_ec$. My question is, why there's a $c$ in the factor?

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It is in cgs units. The magnetic moment of the electron is given by $$\mu_{\rm e}=-g\mu_{\rm B}\frac{\mathbf{S}}{\hbar}.$$ where $g\approx 2$ is the free electron g-factor and $\mu_{\rm B }$ is Bohr magneton. In SI units is written as $\mu_{\rm B}=\frac{\hbar e}{2m_{\rm e}}$ but in cgs it is given by $\mu_{\rm B}^{\rm cgs}=\frac{\hbar e}{2m_{\rm e}c}$.

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  • $\begingroup$ G is not exactly 2.You need to take into account the cloud of virtual particles around the electron. $\endgroup$
    – Jun Seo-He
    Commented Oct 20, 2021 at 15:12
  • $\begingroup$ @JunSeo-He corrected. $\endgroup$
    – Mauricio
    Commented Oct 20, 2021 at 16:07

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