Discovering the Nucleus — Question 3

Question 3

Annular beam geometry yields the Rutherford angular cross section

A parallel beam scatters from a fixed Coulomb center. Use $b=a\cot(\theta/2)$ and axial symmetry: $d\sigma=2\pi b|db|$, $d\Omega=2\pi\sin\theta\,d\theta$.

1. Derive $d\sigma/d\Omega$.

2. Find the ratio of differential cross sections at 30 degrees and 90 degrees, and the scaling when beam energy doubles.

Question 3 - Solution

Annular beam geometry yields the Rutherford angular cross section

(a) Method and calculation

$$\frac{d\sigma}{d\Omega}=\frac{b}{\sin\theta}\left|\frac{db}{d\theta}\right|,\qquad\left|\frac{db}{d\theta}\right|=\frac a2\csc^2(\theta/2).$$$$\boxed{\frac{d\sigma}{d\Omega}=\frac{a^2}{4}\csc^4(\theta/2)=\left(\frac K{4E}\right)^2\csc^4(\theta/2)}.$$

(b) Interpretation and further calculation

$$\boxed{\frac{(d\sigma/d\Omega)_{30}}{(d\sigma/d\Omega)_{90}}=\left(\frac{\sin45^\circ}{\sin15^\circ}\right)^4=55.71}.$$$$a\propto E^{-1}\Rightarrow\boxed{d\sigma/d\Omega\propto E^{-2};\quad E\to2E:\ \sigma\prime/\sigma=1/4}.$$

Physical check

The absolute derivative is essential because b decreases as angle increases. A differential cross section has dimensions of area per solid angle.