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Rutherford scattering

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data

    The differential cross section for Rutherford scattering of alpha particles off a nucleus can be written as:

    [tex] \frac{d\sigma}{d\Omega} = (\frac{zZe^2}{4\pi\epsilon_0})(\frac{1}{4T_\alpha})^2\frac{1}{sin^4 \frac{\theta}{2}} [/tex]

    where z and Z are the atomic numbers of the alpha particle and nucleus respectively, Τα is the initial energy of the alpha particles, and θ is the scattering angle relative to the incoming trajectory.

    Consider an experiment in which a beam of alpha particles is directed at a thin foil of gold. A detector in the form of an annular ring is placed 30 mm downstream of the foil and concentric with the beam direction. The inner radius of the annulus is 5 mm and the outer radius is 7mm. Taking θmin and θmax to be the angular separation between the inner edge and outer edge of the detector relative to the foil find the value of θmin and θmax.

    Now using the differential cross section expression above, and assuming we have azimuthal symmetry so that dΩ=2πsinθdθ, derive an expression for the probability σ for scattering into the detector in terms of (sin(θ/2))^2, θmin and θmax.
    2. Relevant equations

    3. The attempt at a solution

    Let d = 30mm distance from foil to centre of the annulus.
    Rin = 5mm is the inner radius
    Rout = 7mm is the outer radius

    For θmin and θmax we're going to be looking at triangles
    Hypotenuse for inner radius 30.41mm
    Hypotenuse for outer radius 30.81mm

    [tex]\theta_{min} = sin^{-1}(x/H_{min}) = sin^{-1}(5/30.41) = 0.1652 rad [/tex]
    [tex]\theta_{max} = sin^{-1}(x/H_{max}) = sin^{-1}(7/30.81) = 0.2292 rad [/tex]

    I'm stuck on the last part, I assume you take [tex]d\sigma = 2\pi sin\theta d\theta[/tex]
    and replace it in the original equation then integrate θ between θmin and θmax

    [tex] \int d\sigma = \int_{\theta_{min}}^{\theta_{max}}(\frac{zZe^2}{4\pi\epsilon_0})(\frac{1}{4T_\alpha})^2\frac{1}{sin^4 \frac{\theta}{2}}2\pi sin\theta d\theta [/tex]

    but I can't see how I'm going to get an expression involving "derive an expression for the probability σ for scattering into the detector in terms of (sin(θ/2))^2, θmin and θmax."

    Attached Files:

    Last edited: Nov 18, 2008
  2. jcsd
  3. Nov 18, 2008 #2
    Think I've got it.
    \int d\sigma = k\int_{\theta_{min}}^{\theta_{max}}\frac{sin\theta}{sin^4 \frac{\theta}{2}}d\theta = k[\frac{4}{cos \theta - 1} ]_{\theta_{min}}^{\theta_{max}}

    Trig formula: sin^2(x) = 1/2 - 1/2 cos(2x)

    Therefore [tex][\frac{4}{cos \theta - 1} ]_{\theta_{min}}^{\theta_{max}} = [\frac{4}{-2sin^2 {\theta}/2} ]_{\theta_{min}}^{\theta_{max}}[/tex]

    Will the equation still be in terms of (sin(θ/2))^2, if I replace the θ with θmin and θmax?
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