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## Q. 13.2

E. M. Bernstein and colleagues* measured the differential cross section $\sigma(\theta) \text { at the same energy } E_{\alpha}=14.6$ MeV and at the same scattering angle for the $\alpha+{ }^{12} C \rightarrow n+{ }^{15} O$ and $\alpha+{ }^{12} C \rightarrow p+{ }^{15} N$ reactions. They found differential cross sections of 3 mb/sr and 0.2 mb/sr, respectively, for the neutron and proton production at the same scattering angle θ. How much more likely is it that a neutron is produced than a proton?

## Verified Solution

The probability of scattering is simply $n t \sigma$, so let $P_{n} \text { and } P_{p}$ be the neutron and the proton probability, respectively. We also denote the neutron and proton differential cross section at scattering angle $\theta \text { by } \sigma_{n}(\theta)$ and $\sigma_{p}(\theta)$, respectively.

$\frac{P_{n}}{P_{p}}=\frac{n t \sigma_{n}(\theta)}{n t \sigma_{p}(\theta)}=\frac{\sigma_{n}(\theta)}{\sigma_{p}(\theta)}=\frac{3 mb / sr }{0.2 mb / sr }=15$

For the particular scattering angle θ, this ratio is larger than might be expected from the fact that neutrons and protons have similar nuclear interactions. Other factors, such as overcoming the Coulomb barrier and the existence of resonances, are important and will be discussed in Section 13.3.

*E. M. Bernstein et al., Physical Review C 3, 427 (1971).