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Neutrons scattering on Silver plate

  1. May 25, 2014 #1
    1. The problem statement, all variables and given/known data
    A 2 mm thick plate of natural Silver absorbs 11% of neutron flux with kinetic energy 1 eV. What is the total scattering cross section for neutrons?
    ##\rho (Ag)=10500 kg/m^3## and ##M(Ag)=107.9kg/kmol##
    What is the ratio between the calculated cross section and geometric cross section of nucleus?


    2. Relevant equations


    3. The attempt at a solution

    No problems with the first part, but also no idea what to do with the second part of the problem?

    If I am not mistaken, than I should be able to get the geometric cross section from ##\rho (Ag)=10500 kg/m^3## and ##M(Ag)=107.9kg/kmol##; while calculated cross section should be in some relation to scattering?

    Any hints?
     
    Last edited: May 25, 2014
  2. jcsd
  3. May 25, 2014 #2
    Actually, here is what brings me to wrong solution:

    From the first part ##\sigma _{tot}=9,945\cdot 10^{-28}m^{2}=10^{-27}m^{2}##

    Now I said that ##\sigma _{tot}=\pi R^2##. From where ##R_c=(\frac{\sigma _{tot}}{\pi })^{1/2}##. This should be calculated cross section.

    Geometric cross section:

    ##\rho =\frac{m}{V}=\frac{m}{\frac{4}{3}\pi R^3}=\frac{M(Ag)}{\frac{4}{3}\pi R^3N_a}## from where ##R_g=(\frac{3M}{\rho N_a4\pi})^{1/3}##

    Than ##\frac{R_c}{R_g}=0.00011## instead of 11.6.
     
    Last edited: May 25, 2014
  4. May 25, 2014 #3

    SteamKing

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    Is M supposed to be the molar mass of mercury? Mercury has an atomic weight of about 200.6.
     
  5. May 25, 2014 #4
    Ooops, sorry, I made I mistake when posting my question. The element is Silver, not Mercury.

    I will edit my original post.
     
  6. May 26, 2014 #5

    haruspex

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    I very much doubt the two measures should disagree so violently.
    It's hard to say much more because there's so little detail in your post. I've no idea how you compute σtot, so I'll just accept that. For the geometric radius I get roughly 10-10m from your equation. From what I see in Wikipedia, that's about right. Is that what you got? You didn't mix up kmol with mol, right?
     
  7. May 26, 2014 #6
    Ok. My result for this part is the same as the "official".

    Yes, ##R_g=(\frac{3M}{\rho N_a4\pi})^{1/3}=1.6\cdot 10^{-10} m##

    Of course, maybe I understood "geometric" wrong. Maybe something else is meant by geometric and not something you get out of this formula. :/
     
  8. May 26, 2014 #7

    haruspex

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    Just noticed a crucial word... Geometric cross section of NUCLEUS. I believe you computed the cross section of the whole atom.
     
  9. May 26, 2014 #8

    SteamKing

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    The thread title has now been edited to change 'Mercury plate' to 'Silver plate' to avoid further confusion.

    SteamKing
     
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