In the case of a Coulomb potential, V(r)=Z_1Z_2e^2/r, equation (11.70) \frac{d\sigma }{d\Omega }=\left|f(\theta )\right|^2=\frac{4\mu ^2}{\hbar ^4q^2}\left|\int_{0}^{\infty }{r^\prime V(r^\prime )}\sin (qr^\prime )dr^\prime \right| ^2. becomes
\frac{d\sigma }{d\Omega }=\frac{4Z^2_1Z^2_2e^4\mu ^2}{\hbar ^4q^2}\left|\int_{0}^{\infty }{\sin (qr)dr} \right| ^2, (11.75)
where
\int_{0}^{\infty }{\sin (qr)dr}=\underset{\lambda \rightarrow 0}{\lim}\int_{0}^{\infty}{e^{-\lambda r}}\sin (qr)dr=\frac{1}{2i} \underset{\lambda \rightarrow 0}{\lim} \left[ \int_{0}^{\infty}{e^{-(\lambda-iq) r}}dr -\int_{0}^{\infty}{e^{-(\lambda+iq) r}}dr\right]
=\frac{1}{2i}\underset{\lambda \rightarrow 0}{\lim} \left[\frac{1}{\lambda -iq}-\frac{1}{\lambda +iq} \right] =\frac{1}{q}. (11.76)
Now, since q=2k\sin (\theta /2) , an insertion of (11.76) into (11.75) leads to
\frac{d\sigma }{d\Omega }= \left(\frac{2Z_1\mu Z_2e^2}{\hbar ^2q^2} \right) ^2=\left(\frac{Z_1 Z_2\mu e^2}{2\hbar ^2k^2}\right) ^2\sin ^{-4}\left(\frac{\theta }{2} \right) =\frac{Z^2_1Z^2_2e^4}{16E^2}\sin ^{-4}\left(\frac{\theta }{2} \right) , (11.77)
where E=\hbar ^2k^2/2\mu is the kinetic energy of the incident particle. This relation is known as the Rutherford formula or the Coulomb cross section.
(b) (i) Since the mass ratio of the alpha particle to the gold nucleus is roughly equal to the ratio of their atomic masses, m_1/m_2=A_1/A_2=\frac{4}{197}=0.0203, and since \theta _1=60^\circ , equation (11.14) \tan \theta _1=\frac{\sin \theta }{\cos \theta +V_{2_C}/V_{1_C}}=\frac{\sin \theta}{\cos \theta +m_1/m_2}, yields the value of the scattering angle in the CM frame:
\tan 60^\circ =\frac{\sin \theta }{\cos \theta +0.0203} \Longrightarrow \theta =61^ \circ. (11.78)
(ii) The numerical estimate of the cross section can be made easier by rewriting (11.77) in terms of the fine structure constant \alpha =e^2/\hbar c=\frac{1}{137} and \hbar c=197.33 MeV fm:
\frac{d\sigma }{d\Omega }=\frac{Z_1^2Z^2_2}{16E^2} \left(\frac{e^2}{\hbar c} \right)^2(\hbar c)^2 \sin ^{-4}\left(\frac{\theta }{2} \right) =\left(\frac{Z_1Z_2\alpha }{4} \right) ^2\left(\frac{\hbar c}{E} \right) ^2\sin ^{-4}\left(\frac{\theta }{2} \right). (11.79)
Since Z_1=2,Z_2=79,\theta =61^\circ ,\alpha =\frac{1}{137},\hbar c=197.33 MeV fm , and E=8 MeV, we have
\frac{d\sigma }{d\Omega }=\left(\frac{2\times 79}{4\times 137} \right) ^2\left(\frac{197.33MeV fm}{8MeV} \right) ^2\sin ^{-4}(30.5^\circ )
=30.87fm^2=0.31\times 10^{-28}m^2=0.31 barn, (11.80)
where 1 barn =10^{-28}m^2.