Consider the elastic scattering of 50MeV neutrons from a nucleus. The phase shifts measured in this experiment are \delta_{0}=95^{\circ}, \delta_{1}=72^{\circ}, \delta_{2}=60^{\circ}, \delta_{3}=35^{\circ}, \delta_{4}=18^{\circ}, \delta_{5}=5^{\circ} ; all other phase shifts are negligible (i.e., \delta_{l} \simeq 0 for l ≥ 6).
(a) Find the total cross section.
(b) Estimate the radius of the nucleus.
(a) As \delta_{l} \simeq 0 for l ≥ 6 , equation (11.102)
\sigma=\sum_{l=0}^{\infty} \sigma_{l}=\frac{4 \pi}{k^{2}} \sum_{l=0}^{\infty}(2 l+1) \sin ^{2} \delta_{l}, (11.102)
yields
\sigma=\frac{4 \pi}{k^{2}} \sum_{l=0}^{6}(2 l+1) \sin ^{2} \delta_{l}
=\frac{4 \pi}{k^{2}}\left(\sin ^{2} \delta_{0}+3 \sin ^{2} \delta_{1}+5 \sin ^{2} \delta_{2}+7 \sin ^{2} \delta_{3}+9 \sin ^{2} \delta_{4}+11 \sin ^{2} \delta_{5}\right)=\frac{4 \pi}{k^{2}} \times 10.702. (11.178)
To calculate k^{2}, we need simply to use the relation E=\hbar^{2} k^{2} /\left(2 m_{n}\right)=50 MeV, since the neutrons move as free particles before scattering. Using m_{n} c^{2}=939.57 MeV and \hbar c=197.33 MeV fm, we have
k^{2}=\frac{2 m_{n} E}{\hbar^{2}}=\frac{2\left(m_{n} c^{2}\right) E}{(\hbar c)^{2}}=\frac{2(939.57 MeV )(50 MeV )}{(197.33 MeV fm )^{2}}=2.41 fm ^{-2}. (11.179)
An insertion of (11.179) into (11.178) leads to
\sigma=\frac{4 \pi}{2.41 fm ^{-2}} \times 10.702=55.78 fm ^{2}=0.558 \text { barn }. (11.180)
(b) At large values of l, when the neutron is at its closest approach to the nucleus, it feels mainly the effect of the centrifugal potential l(l+1) \hbar^{2} /\left(2 m_{n} r^{2}\right); the effect of the nuclear potential is negligible. We may thus use the approximations E \simeq l(l+1) \hbar^{2} /\left(2 m_{n} r_{c}^{2}\right) \simeq42 \hbar^{2} /\left(2 m_{n} r_{c}^{2}\right) where we have taken l \simeq 6, since \delta_{l} \simeq 0 for l ≥ 6 . A crude value of the radius of the nucleus is then given by
r_{c} \simeq \sqrt{\frac{21 \hbar^{2}}{m_{n} E}}=\sqrt{\frac{21(\hbar c)^{2}}{\left(m_{n} c^{2}\right) E}}=\sqrt{\frac{21(197.33 MeV fm )^{2}}{(939.57 MeV )(50 MeV )}}=4.17 fm. (11.181)