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Nuclear cross section

  1. Mar 16, 2015 #1
    1)
    Alpha particles can react with 48-Ca to produce protons. Consider a beam of alpha particles of current 20nA fired at a think target of Ca of thickness $$1mg/cm^2$$. A detector subtending a solid angle of 4*10^-3 steradians, records 20 protons per second. Determine the total cross section (in mb) for the reaction. State any assumptions made.

    2)

    3)
    So if I=20nA then the initial rate of alphas is:
    $$R_o=\frac{I}{2e}=6.25 \times 10^{10} atoms / sec$$
    And also the number of atoms in the think foil per $m^2$ is $$N=1.25\times 10^{23} atoms/m^2$$.
    But I have no idea what to do from here I understand I'm given rate of initial alphas and rate of detection and a solid angle.
    Could someone just point me in the right direction?
     
    Last edited: Mar 16, 2015
  2. jcsd
  3. Mar 16, 2015 #2

    Orodruin

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    If the detector registers 20 protons per second, how many protons are produced in the target?
    What is the rate of proton production in the target? (Or, more importantly, how does it relate to the cross section?)
     
  4. Mar 16, 2015 #3
    Assuming emission to be a semi-sphere and which is 2pi steradians then the protons per second created would be 31415 per second created so that means that there were that many reactions.
    The only equation I can find is
    $$\sigma = \frac{R}{I} $$
    Reactions per second, over incident particles per second per unit area.
    I know I calculated the number of alpha Per second but I don't quite understand how to get it per unit area and what the the number of 48-Ca atoms has to do with anything. I'm probably missing something simple.
     
  5. Mar 16, 2015 #4
    Oh if i instead workout the number of particles per uni cm^2 and then take the beam to have a area of 1cm^2 ?
     
    Last edited: Mar 16, 2015
  6. Mar 16, 2015 #5

    Orodruin

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    Your final answer should not depend on the area of the beam. I suggest just calling the area A and making sure it cancels out from your final expression.
     
  7. Mar 17, 2015 #6
    I'm really quite stuck, hmm this is what I've done I just don't know how to put it together.

    Let me know if the image can't be read, I will type my workings in latex.
    ImageUploadedByPhysics Forums1426600456.547267.jpg
     
    Last edited: Mar 17, 2015
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