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Homework Help: Hard nuclear physics cross section problem

  1. Nov 13, 2006 #1
    In a scattering experiment, an aluminium foil of thickness L = 10E-6 m is placed in a beam of intensity of 6E12 particles per second. The differential scattering crosss section is known to be of the form.

    [tex]\frac{d\sigma}{d\Omega} = A + Bcos^2 (\theta) [/tex]

    where [tex]\theta [/tex] is the scattering angle and [tex]\Omega [/tex] the solid angle.

    With a detector area of 0.1x0.1 m^2, placed at a distance of 5m from the foil, it is found that the mean counting rate is 40 Hz when [itex]\theta [/itex] is 30 degrees and 31.5Hz when [itex]\theta [/itex] is 60 degrees. Find the values of A and B.


    Clearly I had to set up two simulataneous equations using the different [itex]\theta [/itex] values.
    I've found the number of targets which is just
    [tex] \frac{N_{A}L\rho}{A}[/tex]

    where A is the mass number (27 for Aluminium).

    I looked on wiki and the differential cross section is
    [tex]\frac{d\sigma}{d\Omega} = \frac{scattered flux / unit of solid angle
    }{initial flux / unit of surface}[/tex] but I don't really understand how to relate this to the information I have (I guess the incident flux is just the number of targets x multiplied by the beam intensity per unit area)
     
  2. jcsd
  3. Nov 13, 2006 #2
    Watch out -- The flux is going to be the number of particles that hit the target PER unit area.
     
  4. Nov 13, 2006 #3

    OlderDan

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    To add a bit more to the other reply, I think dσ is proportional to the count rate (counts per second per unit area of the detector), and dΩ is a solid angle that you can calculate from the geometry of the apparatus. I don't think the number of targets enters into this calculation.
     
  5. Nov 13, 2006 #4
    Ok, so the solid angle is just [tex] d\sigma = A / r^2 [/tex] where A is the detector area, and r is the distance between the foil and the detector. I'm still unsure about dσ.

    Just realised the density of aluminium is also quoted in the question, so probably needed.
     
    Last edited: Nov 13, 2006
  6. Nov 13, 2006 #5

    OlderDan

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    The solid angle is dΩ. The total solid angle of a sphere is 4π steradians and the surface area of a sphere is S=4πr² so dΩ = A/S = A/r²
     
  7. Nov 13, 2006 #6
    yup that's what I meant, just put it in wrong my latex seems to take an age to update.
     
  8. Nov 13, 2006 #7
    Think I've worked it out thanks to your hint Dan.
    The scattering fraction
    [tex] \frac{R_s}/{R_i}= \frac{N_{A}L\rho\sigma}{A_m}[/tex]
    [tex] d\sigma = frac{A_mR_s}/{N_AL\rho R_i)} [/tex]
    where R_s = scattered count rate, and R_i is the incident count rate.
     
    Last edited: Nov 13, 2006
  9. Nov 13, 2006 #8

    OlderDan

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    I assume there was more to your problem than just finding the A and B you first asked about.
     
  10. Nov 13, 2006 #9
    no, there isn't I can just say [tex] KR_s = A + Bcos^2 (\theta) [/tex]
    where K is a constant. Not sure why he has including all the blurb.
     
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