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Stern-Gerlack Experiment with silver atoms

  1. Apr 18, 2013 #1
    1. The problem statement, all variables and given/known data

    The magnetic field in a Stern-Gerlach experiment varies along the vertical direction so that the magnetic field increases by 26.00 T each centimeter. The horizontal length of the magnet is 7.1 cm, and the speed of the silver atoms is 922 m/s. The mass of the silver atom is 1.800×10-25 kg. What is the separation of the two silver atom beams as they leave the magnet?

    2. Relevant equations

    [itex] F = - \nabla V [/itex]

    [itex] F_{z} = - m_{l} \mu_{B} \frac{dB_z}{dz} [/itex]

    [itex] B = 2600 z [/itex]

    where z is the vertical position

    3. The attempt at a solution

    I started out thinking I was looking for an angle between the two paths, so I found

    F = -2600

    and [itex] F_z = - \mu_B (2600) [/itex]

    solving for theta:

    [itex]\theta = tan^{-1}(\frac{-2600\mu_{B}}{-2600})[/itex]

    which simplifies to [itex] tan^{-1}(\mu_B)[/itex]

    and that doesn't sit well with me.

    Any ideas?
     
    Last edited: Apr 18, 2013
  2. jcsd
  3. Apr 18, 2013 #2

    Simon Bridge

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    why not?

    why do you have a F as well as Fz?
    how did you use the length of the magnet and the speed of the atoms?
    did you take into account that there is more than one angular momentum state involved?

    note:
    ##F_z = m_{Ag}a_z## isn't it?
     
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