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Stern Gerlach angle?

  1. Aug 26, 2007 #1
    How can you calcualte the angle of the split of the electrons after passing it thorugh a non uniform magnetic field?

    I know how to calculate it by elementary means using the magnitude of S and Sz and doing trig. But is that angle always the same no matter what intensity magnetic field its passed through?
     
    Last edited: Aug 26, 2007
  2. jcsd
  3. Aug 26, 2007 #2

    olgranpappy

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    the actual magnitude of the "split" depends on the details of the apparatus and incoming beam.
     
  4. Aug 26, 2007 #3
    Does it depend on the magnetic quantum number by any chance? It definitely depends on the spin quantum number.
     
  5. Aug 26, 2007 #4

    olgranpappy

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    Yeah, but often one considers atoms that have [tex]\ell=0[/tex], for example Ag atoms (as described in section 1.1 of Sakurai "Modern Quantum Mechanics"), because this type of atom makes for simpler examples.

    Silver can be thought of as having a full d-band and so there is only one "valence" electron in the 5s state. Thus the total orbital angular momentum is zero, but the total spin angular momentum is \hbar/2 (and, of course, the total angular momentum is \hbar/2) and we consider the energy perturbation to be:

    [tex]
    \Delta H \approx \mu_{\textrm{Bohr}}B_0 \sigma_z
    [/tex]

    where [tex]\mu_{\textrm{Bohr}}[/tex] is the Bohr Magneton and B_0 is the external field in the z-direction and \sigma_z is the Pauli matrix. So the atoms shooting out of an "oven" into the apparatus feel a force (only in the region where B_0 is changing--the "fringing" part) due to the changing B_0 field of either plus or minus
    [tex]
    \mu_{\textrm{Bohr}} \frac{d B_0}{dz}
    [/tex]
    since the spin is quantized. And thus there appear two "spots" on the detecting screen.

    I believe that you can figure out the approximate angular distance between the spots using
    [tex]
    2\theta \approx 2\frac{\delta p}{p} = 2\frac{\int F dt}{p} \approx 2\frac{B_0\mu_{\textrm{Bohr}}/v}{mv}
    =\frac{B_0 \mu_{\textrm{Bohr}}}{E_0}
    [/tex]
    where E is the incident energy of the atom. The above is quite approximate indeed and should only hold for \mu B_0 << E_0.
     
    Last edited: Aug 26, 2007
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