108 Chapter 2. Interaction of Radiation with Matter
cases and is also consistent with the experimental observation depicted in Fig.2.4.2.
The classical limit formula is the one that was originally derived by Rutherford.
The total cross section for Rutherford scattering can be determined by integrating
the differential cross section over the whole solid angle Ω.
σtotal=∫
Ωdσ
dΩdΩ (2.4.4)Generally one is interested in predicting the number of particles scattered in a cer-
tain solid angle (see example below). This can be done by simply multiplying the
differential cross section by the element of the solid angle, such as
σ(θ)=dσ
dΩΩ. (2.4.5)
Example:
In his original scattering experiment, Rutherford observed the number of
α-particles scattered off gold nuclei by counting the number of flashes on a
scintillation screen in a dark room. Luckily, with the advent of electronic
detectors, such painstaking work is now not needed. Let us replicate
Rutherford’s experiment but with an electronic detector having a surface
area of 1cm^2. Suppose the source emits 3MeV α-particles with an intensity
of 10^6 s−^1 and the target is a 1μmthick aluminum foil (the setup would be
identical to the one shown in Fig.2.4.1). Assume that the detector can count
individualαparticles with an efficiency of 60%. Compute the number of
counts recorded by the detector at 10^0 and 30^0 relative to the initial direction
of motion of theα-particles. At both angles the distance of the detector from
the interaction point remains 15cm. Take the atomic density of gold to be
6 × 1028 m−^3.Solution:
Let us begin by simplifying equation 2.4.3 for scattering of 3MeV α-particles
by gold nuclei.dσ
dΩ=
[
ZiZte^2
16 π 0 T] 2
1
sin^4 (θ/2)=
[
(2)(79)
(
1. 602 × 10 −^19
) 2
16 π(8. 85 × 10 −^12 )(3× 106 × 1. 602 × 10 −^19 )] 2
1
sin^4 (θ/2)=3. 597 × 10 −^281
sin^4 (θ/2)m^2 /sterHeresterstands for steradian, which is the unit of solid angle. Using this rela-
tion we can compute the differential scattering cross sections for the particles