Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Recombination lines

  1. Mar 2, 2009 #1
    Could someone explain to me what is the "reduced mass" effect on the recombination lines? My teacher told me that would explain why the wavelength I calculated seems a little bit off with the experimental result. I don't seems to find the information anywhere...Thank you in advance.
     
  2. jcsd
  3. Mar 2, 2009 #2
    In whatever analysis you've been explained (Bohr model or Quantum mech), you have used a figure for the kinetic energy term given by p/2m (momentum over twice the mass). This mass is not actually that of the electron, but that of the combined electron nucleus pair, since the nucleus' motion will also contribute slightly (very slightly) to the total energy. The formula for the reduced mass is (m1 x m2)/(m1 + m2), where m1 and m2 are the two masses involved. For hydrogen, this shifts the energy of the lowest state from -13.606 eV to -13.598. More thorough treatments can be found at http://en.wikipedia.org/wiki/Reduced_mass where the origin of the formula is explained.
     
  4. Mar 2, 2009 #3
    Thank you very much ^^. Is that mean now I have to use R_M instead of R_infinity for a "real" Rydberg constant?
     
  5. Mar 2, 2009 #4
    Yeah, that's right. Does that completely sort out the problems with the differences with the spectra. There are other effects which will affect the energy levels, but it depends on the accuracy of your measured spectra whether you want to take them into account. Relativistic effects, spin-orbit coupling, Darwin term, etc.

    Once you start adding in other electrons, you have to take their effects into account as well and things get quite complicated quite fast, with the number of degrees of freedom increasing rapidly.

    But hopefully this correction of the mass will suffice for hydrogenic atoms with your spectra?
     
  6. Mar 2, 2009 #5
    Yeah. I'm doing stuff with Helium spectrum. The weird thing is the yellow line (lambda = 5.87E-7m). I did some research, and it turns out this line is created by neutral helium (He I). If I'm not wrong, neutral Helium should have spectra similar to that of Hydrogen (the electron close to the center cancel the 1 proton charge). But Hydrogen does not have this line. Can this be explained by what you said?
    Thanks ^^
     
  7. Mar 2, 2009 #6

    alxm

    User Avatar
    Science Advisor

    That'd be the Helium 3D line. It's a triplet state (two parallel electron spins), hence the '3'.
    Hydrogen has one electron, so it can only have a doublet state (one unpaired spin). Helium has two electrons and two possible combinations of spin (singlet, triplet). So it has twice the number of energy levels.
    (well, actually it has even more than that. But that effect alone doubles the number)
     
  8. Mar 2, 2009 #7
    I recall in college having to look at line spectra using a ruled grating to measure optical (visible) wavelengths. We looked at line spectra from both hydrogen and deuterium (a hydrogen-like atom with one electron). The lines in hydrogen are very slightly shifted (lowered) in energy due to the larger recoil energy of the proton.
     
  9. Mar 2, 2009 #8
    Thanks guys ^^. One more thing...the line of 504nm and 501nm seems really out of place (created by He I). I found this website, but I cannot understand the "Aki", the "Acc." (Is this frequency?) "configuration", the "terms", the "Ji-Jk", and the "Gi-Gk" (I understand nothing, basically >.<). I can't find them in my book either. Could someone please explain them to me again...? Thank you in advance ^^
    here is the website http://physics.nist.gov/PhysRefData/ASD/lines_form.html
    Enter He I, and the range 500-510 nm, you'll see the result.
     
    Last edited: Mar 2, 2009
  10. Mar 2, 2009 #9

    alxm

    User Avatar
    Science Advisor

    [tex]A_{ki}[/tex] is the transition probability between the states denoted k and i. It's a quantity you can calculate from quantum mechanics, and it's proportional to the intensity of the spectral line.
    Acc.? I guess it's some grading for accuracy of the measurement.
    "configurations" are the electronic states (orbitals) you're transitioning between. The ground state of Helium is the 1s orbital, both electrons in it is 1s2.
    "terms" http://en.wikipedia.org/wiki/Term_symbol" [Broken] for the two states.
    J - the electronic angular momentum of the states.
    g - not sure. Landé g-factor?
     
    Last edited by a moderator: May 4, 2017
  11. Mar 2, 2009 #10
    Thank you ^^
     
  12. Mar 4, 2009 #11
    Oh! Could someone please also explain to me what is E, and why cm-1 as unit?
    For the transition, there's something I don't understand. By writing 1s2s-1s3p, does it means the electron goes from 2s to 3p? But isn't that going up (further from the nucleus)? How is that gonna emit photon then...?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Recombination lines
  1. Balmer Lines (Replies: 4)

Loading...