Discovering the Nucleus — Question 2

Question 2

Impact parameter controls the Coulomb scattering angle

For repulsive scattering from a fixed point nucleus, take $b=a\cot(\theta/2)$, where $a=K/(2E)$ and $K=Z_1Z_2e^2/(4\pi\varepsilon_0)$. Let $a=10.0\,\mathrm{fm}$.

1. Find the impact parameters for scattering through 60 degrees and 120 degrees.

2. For a fixed impact parameter $b=10.0\,\mathrm{fm}$, find the initial angle and the angle after E is doubled.

Question 2 - Solution

Impact parameter controls the Coulomb scattering angle

(a) Method and calculation

$$\boxed{b_{60}=10\cot30^\circ=17.32\,\mathrm{fm}}.$$$$\boxed{b_{120}=10\cot60^\circ=5.774\,\mathrm{fm}}.$$

(b) Interpretation and further calculation

$$\tan(\theta/2)=a/b=1\Rightarrow\boxed{\theta=90.0^\circ}.$$$$E\to2E\Rightarrow a\to a/2\Rightarrow\boxed{\theta\prime=2\tan^{-1}(.5)=53.13^\circ}.$$

Physical check

Smaller impact parameters probe stronger deflections. At a fixed trajectory offset, a faster projectile spends less time near the target and bends less.