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Stern-gerlach experiment

  1. May 13, 2006 #1
    OK I just have one more question haha.

    Consider the original Stern-Gerlach experiment employing an atomic beam of silver, for which the magnetic moment is due entirely to the spin of the single valence electron of the silver atom. Assuming the magnetic field B has a magnitude 0.500 T, compute the energy difference in electron volts of the silver atoms in the two exiting beams.

    What I've done:

    So we're looking at an electron, which has a spin quantum number of s=1/2.
    So the z component of the spin angular momentum will be.

    Sz = ms*hbar

    Where ms is the spin magnetic quantum number. There are two different orientations of 1/2 and -1/2 that result in the different defelctions of the beams.

    So the energy for ms = 1/2 is:

    E = (1/2)*hbar*(e/2m)*g*B

    And for ms = -1/2

    E = (-1/2)*hbar*(e/2m)*g*B

    To find the energy difference I subtracted the two energies, which yielded:

    deltaE = hbar*(e/2m)*g*B

    e is the charge of the electron and m is the mass of the electron and g is the g factor (g = 2.00232).

    So the above equation can now be expressed as

    deltaE = Ub*g*B

    Where Ub is the bohr magneton.

    But this answer is wrong. It would work out if I didn't use the refined value of g and just used 2.

    What am I doing wrong?
  2. jcsd
  3. May 13, 2006 #2


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    I am not quite sure what you mean. Do you mean that you would get the correct answer if you used g =2 instead of 2.00232?

    If this what you mean than two comments: a) if you work within three sig figs, it does not make any difference and b) most elementary books use g=2, i.e. they neglect vacuum fluctuations effects.

    So if this is your only problem, I would say to use g=2 and not worry about the difference of one part in a thousand
  4. May 13, 2006 #3
    yeah that was my only problem. I just wasn't sure if the book wants you to use g = 2.00232 or not. But yeah I'll just stick with using 2 then.
    Cheers for that.
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