I have the following data: the separation between two zeeman levels is 6.275 GHz in a magnetic field of 4.6 T.

Calculate e/m (m is the electron mass) and distinguish if is normal or anomalous Zeeman effect.

I find that $$ e/m = \delta\nu \,4\pi/B $$

But how to determine which one Zeeman is?


The Zeeman effect gives a frequency shift of $$ h\,\Delta \nu = g\mu_B B\,\Delta m, $$ under a magnetic field $B$, where $\mu_B=e\hbar/2m_e=9.3\times10^{-24}\:\mathrm{J/T}$ is the Bohr magneton, $g$ is the gyromagnetic ratio, and $\Delta m=1$ is the change in magnetic quantum number.

The normal Zeeman effect is caused by the magnetic dipole moment of the orbital motion of the electron, which has gyromagnetic ratio $g=1$. The anomalous Zeeman effect (to the extent that the term is used at all in the modern literature) refers to the shifts coming from the spin component, which has gyromagnetic ratio $g\approx 2$.

In your case, we have $$ g =\frac{h\,\Delta \nu}{\mu_B B} = \frac{h\times 6.275 \:\mathrm{GHz}}{\mu_B\times 4.6\:\mathrm{T}} \approx0.097, $$ which makes no sense. At this stage, and without more context, there's little to say beyond: double-check your numbers.


Normal Zeeman effect occurs in strong magnetic field but anomalous Zeeman effect occurs in weak magnetic field

  • 1
    $\begingroup$ That's not what that distinction is. You're thinking of the Paschen-Back effect (i.e. the Zeeman effect at high field strengths). The anomalous Zeeman effect refers to shifts coming from the spin of the electrons. $\endgroup$ – Emilio Pisanty Mar 27 '17 at 11:28

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