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Absorption length

  1. Jan 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Aluminium (Ar = 26.98, ρ = 2700 kg/m3) has cross-section of thermal neutron absorption 0.23 barn. What is the absorption length of thermal neutrons in the aluminium? What value decreases neutron intensity in 0.6 mm, 8 mm and 80 mm aluminium?

    2. Relevant equations
    Beer-Lambert Law

    d0f3f6ed693df5ca4b1a717a95749ac8.png
    I'm not sure what x and lambda are in this equation.

    3. The attempt at a solution
    I'm not sure how to calculate the absorption length from the information provided.

    Absorption length
    [PLAIN]http://gisaxs.com/images/math/c/5/0/c50b9e82e318d4c163e4b1b060f7daf5.png, [Broken] the distance over which the intensity falls to 1/e.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 4, 2016 #2

    BvU

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    is strange when you then state
    . Or are you worried about the relation between ##P## and intensity ?
    Is it really that hard to imagine what ##x## could be ?
    Did you Google some examples ?
     
    Last edited by a moderator: May 7, 2017
  4. Jan 5, 2016 #3
    The intensity falls to 1/e = absorption length. But what is the original intensity?
     
  5. Jan 5, 2016 #4

    BvU

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    Why not use 1 for the original intensity ? Or ##I_0## and gamble that ##P = {I\over I_0} ## :smile: ? At least then you can move on !
     
  6. Jan 5, 2016 #5
    ##P = {I\over I_0} ## = 1

    1/e = .367
    1-.367879 = 0.633

    The intensity has fallen, I'd assume x is the distance the thermal neutron has traveled through the atom, maybe. not sure about lambda though.
     
  7. Jan 5, 2016 #6

    BvU

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    We're not making much headway like this. Can you imagine an aluminum plate with a certain thickness, for example 0.6 mm, at which a beam of thermal neutrons is directed. The plate is a void with an awful lot of aluminum atoms. For a neutron in the beam this looks like a volume with a certain density of targets with an area of 0.23 barn per atom, the neutron cross section.
    Thicker plate, more targets. Your ##\lambda## has something to do with the density and the area of these targets.

    You have a textbook or lecture notes to introduce you to the matter at hand ? This exercise doesn't come out of the blue, I should think?
     
  8. Jan 5, 2016 #7
    You'd be surprised...

    λ = density * area of sphere ? (Assuming nucleus has a sphere shape)
    = density * 4*pi*radius of atomic nucleus
     
  9. Jan 5, 2016 #8
    If I take the value of kinetic energy from that link you posted. 0.025 eV I can calculated the scattering cross section. From there I can calculate the absorption cross section.
     
  10. Jan 5, 2016 #9

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    My impression is the 0.23 nb is the absorption cross section. The exercise is a lot simpler than you now make it look.
     
  11. Jan 5, 2016 #10
    Yes, that makes sense. 0.23 is the absorption cross section. I have absolutely nothing in my lecture slides relating to neutron cross section or neutron absorption. No mathematical equation in which I can understand what happens. Is λ the mean free path?
     
  12. Jan 5, 2016 #11
    Σ=attenuation coefficient cm-1
    N= Atom density
    σ = cross section

    Σ=Nσ
    Σ=621 cm-1

    λ=1/Σ= 1/621=0.00161
     
  13. Jan 5, 2016 #12

    BvU

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    What do you get for N ? And how come your units end up as cm-1 ?
     
  14. Jan 5, 2016 #13
    Sorry, N = density = ρ = 2700 kg/m3

    I can see this isn't correct now though through dimensional analysis. density * cross section = kg/m3 * m2 ≠ cm-1
     
  15. Jan 5, 2016 #14

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    You want the number of Al atoms per volume, not the density in kilograms per volume !
     
  16. Jan 6, 2016 #15
    oh ok, so number density = number of objects / volume = 100.74

    100.74*.23 = 23.017 cm-1
    λ=1/Σ= 1/23.017=0.043446

    0.367 = e^-x/lambda
    -1.00239 = -x/0.043446 =
    x= -0.04355

    This answer is still incorrect. The solution is 0.639 m according to my textbook but I don't see how I can get that.
     
  17. Jan 6, 2016 #16

    BvU

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    Do you seriously believe there are 100 Al atoms in a 1 m3 block of Aluminum ? Or do you work in some other system of units ?
     
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