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Homework Help: Nuclear Cross Section

  1. Aug 17, 2014 #1
    1. The problem statement, all variables and given/known data

    What is the cross section if the density of atoms in a material is 10^26 metres-3, and 0.04% of a beam of neutrons is stopped in a 2 cm slab of the material ?

    2. Relevant equations

    Sigma = event rate per nuclei / incident flux

    3. The attempt at a solution

    I have no idea ..
  2. jcsd
  3. Aug 17, 2014 #2


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    If the density would be 1/meter^3 and 1% would be stopped in 1meter (imagine a cube of 1m side length with exactly one atom in it), can you calculate the cross-section?
    Then you just have to scale this to get the original numbers.
  4. Aug 17, 2014 #3
    Cant really visualise this ... but I guess there would 10^26m-3 with 0.04 % stopped in it ? I need to calculate the barns. Still not sure about the 2 cm mentioned ...

    I have gone away and come up with the following -

    I have converted the 10^26 m-3 to centimetres = 1.0000^32cm3

    I have then calculated using a formula found on the internet


    1.0000^32cm3 x 2cm x 6.022^23 / 0.04 = 3.00 x 10^57

    I cannot see where I am wrong but some advice would be a big help

    I am assuming you must convert the m-3 to cm3, and calculated my answer as per the research article I found. Thank you
    Last edited: Aug 17, 2014
  5. Aug 17, 2014 #4


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    Can one write the relationship between the macroscopic cross-section, ∑, and the microscopic cross-section, σ, which is what one wishes to find?

    Then write the equation for attenuation of particles passing through a slab of material, i.e., write the equation for dI(x)/dx, where I(x) is the beam intensity as a function of distance from the surface into the material.
  6. Aug 17, 2014 #5


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    The formula uses the density in g/cm^3 and uses the Avogadro constant to convert this to atoms/cm^3. You know atoms/cm^3 so you can skip that step in the formula.

    This is easier to see if you work with units everywhere (a good habit in general).
  7. Aug 17, 2014 #6
    Thank you for your help ..

    0.04 % / 1.0000^32cm3 x 2cm = 8.00 x 10^-34

    that is N events (0.04%) / beam particles per unit area (1.0000^32cm3 / target particles (2cm)
  8. Aug 17, 2014 #7


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    The units are missing, but apart from that it looks good.
  9. Aug 17, 2014 #8
    Thanks struggled to solve that you have both been very helpful .. I appreciate your time and trouble so much ..
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