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Energy splitting in hyperfine structure

  1. Oct 23, 2016 #1
    1. The problem statement, all variables and given/known data
    I've been given the worked answer to a very similar question, but there's a step I don't understand so I can't apply it. My question asks:
    An energy level with ##J=\frac{11}{2}## has six hyperfine sub-levels with these relative energies:
    0, 2.51, 4.71, 6.59, 8.16, 9.41 (GHz)
    What is the value of I? Show the spacing obeys the interval rule and determine the value of F for each sub-level.

    2. Relevant equations


    3. The attempt at a solution
    So based on this example question I have, start from the fact that ##F = |I-J|....(I+J)##, integer steps, and label the lowest sub-level as ##F_0 = |I-\frac{11}{2}|##. That means I have energy levels from ##F_0## to ##F_0 +5##. Then I can make a table:
    ##
    \begin{array}{|c|c|}
    \hline
    F & \Delta E_F -\Delta E_{F-1}\\ \hline
    F_0 +5 & 1.25 \\ \hline
    F_0 + 4 & 1.57 \\ \hline
    F_0 + 3 & 1.88 \\ \hline
    F_0 + 2 & 2.2 \\ \hline
    F_0 + 1 & 2.51 \\ \hline
    \end{array}##
    Found by referring back to the listed energy levels, so ##F_0 + 5 = 9.41 - 8.16## and so on. Then take ratios of energy splittings, and here's what I don't understand. From
    ##\frac{F_0+5}{F_0 + 4} = \frac{1.25}{1.57}##, a value for ##F_0 +5## is deduced.

    In the example it states
    ##\frac{F_0 +2}{F_0 +1} = \frac{202}{151}## therefore ##F_0+1 = 3##. How did they work that out? They have ##F_0 = |I - \frac{3}{2}|## and the energy level ##F_0## is at ##151##MeV. Don't understand that step. Thanks for any help!
     
  2. jcsd
  3. Oct 23, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    By finding a small fraction that fits, in this case ##\frac{202}{151} \approx \frac 4 3##. You can find a similar fraction if you divide 1.57 by 1.25, and note a pattern if you continue with 1.88/1.57 and so on.

    151 MeV is way too large for the hyperfine structure.
     
  4. Oct 23, 2016 #3
    Apologies, I meant MHz! 151MeV would be pretty huge. I don't think I can edit it anymore.
     
  5. Oct 24, 2016 #4
    Well for my question ##\frac{F_0+5}{F_0+4} = \frac{1.25}{1.57} \approx 0.796 \approx \frac{4}{5}## which gives ##F_0 = -1## or ##1##, so I think I'm doing something wrong. And looking at the next fraction,
    ##\frac{F_0+4}{F_0+3} = \frac{1.57}{1.88} \approx \frac{5}{6}## again gives two different answers for ##F_0## of 1 and 2. What am I doing wrong? Are the fractions I've approximated to incorrect?
     
  6. Oct 24, 2016 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    It has been a long time since I last worked with hyperfine structures, but the series of fractions is so clear that it must be the right approach.
     
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