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Scattering problem

  1. Apr 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A beam of particles strikes a wall containing 2 × 10^29 atoms/m^3. Each atom behaves as a hard sphere of radius 3 × 10^–15 m.

    Find the thickness of the wall such that exactly half the incident particles go through without scattering.

    2. Relevant equations

    N(sc) = N(inc) × n(tar) × σ ---> (1)
    N(sc) = the number of scattered particles
    N(inc) = the number of incident particles
    n(tar) = the target density
    σ = cross sectional area of the target = πR^2
    n(tar) = ρt/m ---> (2)
    ρ = density
    t = thickness of the wall
    m = mass

    3. The attempt at a solution

    Hello everyone; this is a straightforward scattering problem but I seem to be missing something here. This is what I did so far:
    From the given information I have:
    N(sc) = ½ N(inc) ---> 2N(sc) = N(inc)
    R = 3 × 10^–15 m ---> σ = πR^2 = 9π × 10^–30 m^2
    Plugging these values in equation (1), I can get the value of n(tar) which turned out to be 1.77 × 10^28 m^–2
    Having this value of n(tar), I can substitute into equation (2) and solve for t. This is where I’m stuck, as I’m supposed to find numerical values for ρ and m in order for me to find t. These values can be obtained from the given information that the wall contains 2 × 10^29 atoms/m^3, and I think that I can get my values for ρ and m from the following equation:
    The number of atoms per unit volume = (Avogadro’s number × density)/atomic mass
    That is: 2 × 10^29 atoms/m^3 = (6.022 × 10^23 mol^–1 × density)/atomic mass
    This is where I’m not sure how to proceed, as I need to find the density and atomic mass (and convert it to kg as well) from this equation; any help would be appreciated.
     
  2. jcsd
  3. Apr 10, 2009 #2

    Astronuc

    User Avatar

    Staff: Mentor

    Be careful here.

    We know that the macroscopic cross-section, Σ, for an interaction is given by Σ = nσ, where n is the atomic density and σ is the microscopic cross section.

    So Σs = nσs.

    Now we also know that N(x) = No exp (-Σs x), where x is the distance traveled, and No is the initial particle intensity (we could use I(x) and Io).

    All we need to do is find the distance (thickness) t at which half of the particles reach without scattering, and so they leave the other side (of slab thickness t), without having scattered.

    So what is N(t)/No?
     
  4. Apr 11, 2009 #3
    Now it makes sense using your equations; still straightforward, but different way of looking at things. Thank you very much for your help Astronuc, I really appreciate it.
     
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