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Topic 11 of 11

Classical Scattering

Scattering experiments are how physicists probe the forces between particles. In classical mechanics, a beam of particles with impact parameter $b$ is deflected by angle $\Theta(b)$. The differential cross section $d\sigma/d\Omega$ tells you the rate at which particles are scattered into each solid angle.

Key Concepts

Impact Parameter

b

is the perpendicular distance from the beam axis to the (unperturbed) trajectory of a particle. For a repulsive force, larger

b

gives smaller deflection;

b=0

gives

\Theta=\pi

(head-on backscatter).

Deflection Function

\Theta(b)

maps impact parameter to scattering angle. For Coulomb scattering:

\cot(\Theta/2)=b/a

where

a=kZz e^2/(2E_{ m cm})

Differential Cross Section

d\sigma/d\Omega=b/(\sin\Theta)|db/d\Theta|

. It has units of area per steradian and measures how effectively the target scatters particles into angle

d\Omega

Total Cross Section

\sigma_{\rm tot}=\int(d\sigma/d\Omega)\,d\Omega

. For hard-sphere scattering:

\sigma_{\rm tot}=\pi R^2

, the geometric cross section. Coulomb scattering has infinite total cross section.

Key Equations

Rutherford formula

\frac{d\sigma}{d\Omega} = \left(\frac{a}{4\sin^2(\Theta/2)}\right)^2

Differential cross section for Coulomb scattering; a = kZze²/(2E).

Hard sphere

b = R\cos(\Theta/2), \quad \frac{d\sigma}{d\Omega} = \frac{R^2}{4}

Uniform differential cross section; total cross section = πR².

Total cross section

\sigma_{\rm tot} = \int \frac{d\sigma}{d\Omega}\,d\Omega = 2\pi\int_0^\pi \frac{d\sigma}{d\Omega}\sin\Theta\,d\Theta

Integral over all solid angles; for hard sphere, σ = πR².

Worked Example

Hard Sphere Total Cross Section

Problem

Particles scatter off a hard sphere of radius $R=2.0$ fm. Find: (a) the differential cross section and (b) the total cross section.

Solution

(a) For a hard sphere, $d\sigma/d\Omega=R^2/4$ — uniform over all angles.

\frac{d\sigma}{d\Omega} = \frac{(2.0)^2}{4} = 1.0\text{ fm}^2/\text{sr}

(b) Integrate over the full solid angle $4\pi$ sr:

\sigma_{\rm tot} = \frac{R^2}{4}\times4\pi = \pi R^2 = \pi(2.0)^2 = 4\pi \approx 12.6\text{ fm}^2

Answer

d\sigma/d\Omega=1.0

fm²/sr;

\sigma_{\rm tot}=\pi R^2=12.6

fm².

Practice

Exercises

7 problems

1 of 7

A hard sphere has radius $R=3.0$ fm. Find the differential cross section $d\sigma/d\Omega$ (in fm²/sr).

Your answer

fm²/sr

2 of 7

Same hard sphere ( $R=3.0$ fm). Find the total cross section $\sigma_{\rm tot}$ (in fm²).

Your answer

fm²

3 of 7

In Coulomb scattering, the distance of closest approach for head-on ( $b=0$ ) collision is $d=kZze^2/E$ . For $Z=z=1$ , $k=8.99\times10^9$ N·m²/C², $E=1.0\times10^{-13}$ J, find $d$ (in fm). ( $e=1.6\times10^{-19}$ C, 1 fm = $10^{-15}$ m.)

Unlock Exercise 3