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Zeeman effect problem

  1. Oct 29, 2012 #1
    1. The problem statement, all variables and given/known data

    "The normal Zeeman effect splits a spectral line at frequency [itex]\nu_{0}[/itex] and two satellite lines at [itex]\nu_{0}[/itex] ± [itex]eB/(4\pi m_{e})[/itex]. By what amount (in angstroms) are the satellite lines of the hydrogen Balmer [itex]\alpha[/itex] line ([itex]\lambda_{0} = 6562.81 Å[/itex]) split from the central component in a typical sunspot?

    Given value for B in a sunspot: .1 T

    2. Relevant equations

    [itex]\lambda[/itex] = c/[itex]\nu[/itex]
    d[itex]\lambda[/itex] = c d/d[itex]\nu[/itex]

    3. The attempt at a solution

    I've tried plugging things into [itex]eB/(4\pi m_{e})[/itex] to find the change in frequency for the satellite lines, but I'm not getting a value in hertz, so I'm not exactly sure what I'm doing wrong. I'm pretty sure that once I get an actual frequency out of this I can just use the relevant equations to find the difference in wavelength.
     
  2. jcsd
  3. Oct 29, 2012 #2
    Did you calculate the individual wavelengths and take their difference? or did you calculate the difference in frequency and calculate a wavelength with that?
     
  4. Oct 29, 2012 #3
    If I calculate the frequency based off the given wavelength, I get 4.56*10^14 hertz, but that doesn't get me any further I don't think.

    The problem is the difference between the initial frequency and the satellite lines. I can't add or subtract the difference because the difference isn't a frequency, it's some nonsense units (seconds^-2 ampere^-1).
     
  5. Oct 29, 2012 #4
    Try writing out the units of each piece in the SI base units. See what happens.
     
  6. Oct 29, 2012 #5
    I think this is maybe where I'm missing something?

    A tesla divided by an electron mass is giving me 1 per second per ampere, and wolfram seems to agree with this.

    [itex]\nu_{0} + s^{-1}A^{-1} = Hz + s^{-1}A^{-1}[/itex] is adding incompatible units, so I'm pretty sure I can't do it.
     
  7. Oct 29, 2012 #6
    What about the electron charge?
     
  8. Oct 29, 2012 #7
    Also, I'm getting per second squared in Wolfram.
     
  9. Oct 29, 2012 #8
    Oh okay I see now. I was so sure e was the base of the natural log. I actually tried looking around for other things it could stand for, but putting it in as the electron charge on wolfram made it work.

    Thanks!
     
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