Discovering the Nucleus — Question 4

Question 4

Integrating over a backward cone avoids a small-angle divergence

Use $b=a\cot(\theta/2)$ with $a=12.0\,\mathrm{fm}$. Every trajectory with $b<b_0$ scatters through an angle greater than $\theta_0$.

1. Find the integrated cross section for $\theta>\theta_0$ and evaluate it at $\theta_0=90^\circ$. Use $1\,\mathrm{barn}=100\,\mathrm{fm^2}$.

2. Find the cross section for $60^\circ<\theta<120^\circ$.

Question 4 - Solution

Integrating over a backward cone avoids a small-angle divergence

(a) Method and calculation

$$\sigma(\theta>\theta_0)=\int_0^{b_0}2\pi b\,db=\boxed{\pi a^2\cot^2(\theta_0/2)}.$$$$\boxed{\sigma(\theta>90^\circ)=144\pi\,\mathrm{fm^2}=4.524\,\mathrm{barn}}.$$

(b) Interpretation and further calculation

$$\sigma_{60\text{--}120}=\sigma(>60^\circ)-\sigma(>120^\circ)=\pi a^2(3-1/3).$$$$\boxed{\sigma_{60\text{--}120}=384\pi\,\mathrm{fm^2}=12.06\,\mathrm{barn}}.$$

Physical check

A finite angular cutoff produces a finite result. The ideal unscreened Coulomb total cross section diverges as the cutoff tends to zero.