1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Zeeman effect physics 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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Zeeman effect physics problem
  1. Zeeman effect (Replies: 4)

  2. Zeeman Effect (Replies: 1)

  3. Zeeman effect problem (Replies: 3)

  4. Zeeman Effect (Replies: 18)

  5. Zeeman Effect (Replies: 1)