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Homework Help: Half-value thickness for barium 137m

  1. May 6, 2017 #1
    1. The problem statement, all variables and given/known data
    The half-value layer (thickness) of gamma rays in lead is dependent on the energy of the gamma photons before they enter the lead. Find the graph which shows the relationship between the energy of the gamma photons (pasted below) and the half-value layer and read their energy. In your data book (we didn't get one or even a photocopy of the page), look up the energy of gamma photons emitted by barium 137* and compare it with the energy from the graph.
    Barium 137 hvl over meV copy.png
    2. Relevant equations

    3. The attempt at a solution
    Looking at the graph, it seems to be telling me that an increase in gamma energy at the entry point of a material corresponds with an increase in material thickness needed to cut this energy by half. I'm not really sure I understand the second part of the question, though. For one thing I don't have any data for the gamma energy associated with barium 137*, though I did read (it may have been on Wikipedia) that gamma energies are typically a few hundred kilo-electronvolts. For another, assuming I did manage to find the energy, what does it mean by 'compare it with the energy from the graph'? I imagine it means that I would take the stated energy and insert it into upload_2017-5-6_10-26-28.png
    as a function I(x) and then solve for I(0) and I(n), comparing the results with several points on the graph shown. I hope I am on the right track. Could someone please tell me what the gamma energy for barium 137* is?
  2. jcsd
  3. May 6, 2017 #2


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    How about a web search on "barium 137"?
  4. May 6, 2017 #3
    Hi kuruman,
    I conducted several, but got sick of pages and pages of information about half-life etc. Luckily I've just tried again and lo and behold I found a page which told me the following: 'The energies of both the beta decay of cesium-137 and the subsequent gamma decay of the excited barium 137 are 512 keV and 662 keV, respectively.', which I'm going to run with.
    Thank you.
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