1. Not finding help here? Sign up for a free 30min 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!

Calculate the change in energy vs. B for each state separately?

  1. Aug 16, 2008 #1
    Over a He-lamp a weak vertical magnetic field ( B = 300mT) is beeing applied. The light from the lamp is beeing studies with a high resolution spectrometer in the direction of the B-field. What will the wavelength difference between the observed Zeeman components be in the transmission 1s2p 1P - 1s3s 1S at 728.13nm

    I realize that since we look in the direction of the B-field I understand that we'll only be able to see sigma-transmissions (delta M_j = -1 or +1). I've also calculated that g_j will be 1 for 1P and 3/2 for 1S.

    But from there I dont really know how to continue.
     
  2. jcsd
  3. Aug 16, 2008 #2

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Re: Zeemancomponents

    Can you calculate the change in energy vs. B for each state separately?
     
  4. Aug 16, 2008 #3
    Re: Zeemancomponents

    You tell me.
    When I look in my formula sheet I find.

    "For weak fields, hfs:
    E_ZE = g_F * my_B * B * M_F"

    I suppose that E_ZE is the energy separation? But I dont see how I can calculate g_F, since it contains the nuclear spin I, which I dont have. And I'm not even sure if thats the right way to go. I'm kind of calculating in the dark here.
     
  5. Aug 16, 2008 #4

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Re: Zeemancomponents

    E_ZE is the shift in energy of a particular level.

    Helium has zero nuclear spin, so there is no hfs. There is fine structure however. There should be a similar formula involving g_J and m_J.
     
  6. Aug 17, 2008 #5
    Re: Zeemancomponents

    Ok, this starts to make sence. I figured that S only have M_j = 0, which means that the're only 2 transmissions, so I calculated the difference to E_ZE(M_j = 1) - E_ZE(M_j = -1) and got delta f = 8.398GHz, which I've got confirmed is right.

    but they ask for the wavelength difference, so there's probably some more work to do. If I didnt know that this gives me the wrong answear I would just go with lambda = c/(delta f). But that doesnt seem right. Why?

    they say that they observe the transmission at 728.13nm, what does that really mean? The solution states the following.
    lambda = c/f => delta lambda = (-) lambda^2/c * delta f.

    First of all I dont understand why they do that delta lambda calculation, and when I try to get to the same conclusion I go like this
    delta lambda = c/f_1 - c/f_2 = cf_2/(f_1 * f_2) - cf_1/(f_1*f_2) = c*delta f * 1/(f_1*f_2). And the only way I can connect this to the result they show in the solution is if c^2/(f_1*f_2) = lambda, but why would it? And why does the lambda from the text come in here?

    A lot of blanks still remain.
     
  7. Aug 17, 2008 #6

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Re: Zeemancomponents

    Good, that's reassuring.

    Because lambda is really c/f, not c/(delta f). That is the basis for this calculation, along with:

    [tex]
    \Delta \lambda = \lambda_1 - \lambda_2 = \frac{c}{f_1} -\frac{c}{f_2}
    [/tex]

    as you use in your derivation below.

    The transition (not transmission) between the two states occurs at a wavelength of 728.13nm. Therefore the spectrum of light from the lamp contains this wavelength, and it is observed in the spectrometer.

    So far so good ...

    It's actually lambda^2, not lambda. It's an approximation.

    Another useful piece of information is that the change in wavelength is a small fraction of the actual wavelength. Likewise, the change in frequency is a small fraction of the actual frequency. You can verify this be comparing your calculated frequency change of 8.398 GHz to the actual frequency of c/728.13nm.
    What that does is make f1 and f2 nearly equal to each other and to c/lambda, where lambda is the value given in the problem statement. You can use that to get your expression in terms of lambda:

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

Have something to add?