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Question about nuclear cross sections

  1. Oct 16, 2004 #1
    This is a small part of a homework problem in which I need to calculate neutron flux. I have the neutron beam hitting a copper target, and two reactions take place Co-59(n,p)Co60metastable (cross section=19barn) and Co-59(n,p)Co60ground (cs=18barn). What is the cross section overall? Do I simply say there is a 18+19=37 barn cross section overall? I'm not so sure, it makes sense to add them thinking about it from a probability standpoint, but from an area of target standpoint, I think it should be somewhere between 18 and 19. Any thoughts? Thanks.
  2. jcsd
  3. Oct 16, 2004 #2


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    It is simply the sum, almost by definition. The first cross section is a normalized way of giving the probability that a neutron plus Co59 will give you a metastable Co60. The second one is the probability that the neutron plus Co59 will give you a ground state Co60. These events are mutually exclusive of course, so the probability of obtaining Co60 (metastable or ground state) is the sum of both. In fact, the cross sections have been SPLIT UP in 18 and 19: the total cross section Co59->Co60 is 37 barn, and that has been refined in a fraction metastable and another fraction groundstate.
    I think I see where you are having an issue, in that "the surface taken to interact with the neutron is 18 for one interaction, and 19 for another one, so I'd have a weighted surface for both". No, think of the Co59 nucleus as having a surface of 37 barn, and if you hit the upper 19 barn of it, you'll have a metastable state, and if you hit the lower 18 barn, you'll have a ground state. But then be carefull with this picture :smile: After all, it is a PICTURE, because cross sections are nothing else but normalized probabilities.

  4. Oct 16, 2004 #3
    Thanks so much, you reply helped me see what I had suspected. Now on to bigger and more confusing problems...
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