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Find most likely values of L, S, J for barium

  1. Nov 2, 2016 #1
    1. The problem statement, all variables and given/known data
    The spectrum of barium contains the following series of energy levels (given relative to ground):
    1.521, 1.567, 1.676 (all in eV)
    What are the most likely values of L, S, J?

    2. Relevant equations


    3. The attempt at a solution
    I tried using the Lande interval rule, giving the lowest state the value j, then the 1.567 level j+1 and j+2 for 1.676. If the change between 1.521 and 1.567 is ##\Delta E = 0.046## and the change between 1.567 and 1.676 is ##\Delta E_2 = 0.109##, then ##\Delta E_2 \approx \frac{12}{5} \Delta E##. So by the interval rule
    ##\Delta E = 2B(j+1)## (1)
    ##\frac{12}{5}\Delta E = 2B(j+2)## (2)
    Dividing (1) by (2) to get rid of the constant,
    ##\frac{5}{12} = \frac{j+1}{j+2}## which would give me ##j = -\frac{2}{7}##. Which is wrong. But that's the only method I can think of! Have I made a mistake or should I be using a different approach?

    I'm also very confused by the phrase 'most likely' values; should the method I'm meant to use give multiple possibilities? Thanks for any help!
     
  2. jcsd
  3. Nov 7, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Nov 8, 2016 #3

    DrDu

    User Avatar
    Science Advisor

    Do you know other relevant rules beside Lande? Even then, it is a rule, not an exact law. What values for j are possible in Barium (e.g. integer, half integer ...)? What is the nearest possible value to the one you calculated?
     
  5. Nov 8, 2016 #4
    Hi, thanks for your reply! Fairly sure I've solved it, assumed the levels were part of a triplet as the question says a series of energy levels, which would mean the j values are j, j+1, j+2 with j associated with the lowest energy levels. Then S = 1 for a triplet and J = L + S gives you the L value.
     
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