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## Q. 13.4

In Example 13.2 we noted that the ${ }^{12} C (\alpha, n)^{15} O$ reaction cross section was much larger than that of the ${ }^{12} C (\alpha, p)^{15} N$ reaction at $E_{\alpha}=14.6$ MeV. To what final excitation energy is ${ }^{16} O$* excited in this reaction?

Strategy We need to find the sum of two energies. The first is the kinetic energy of mass [Equation (13.12)]. The second is the excitation energy $E\left({ }^{16} O ^{*}\right)$ due to just the masses when an α particle and ${ }^{12} C$ form to make ${ }^{16} O$ [Equation (13.11)].

$E\left( CN ^{*}\right)=M_{x} c^{2}+M_{X} c^{2}-M_{C N} c^{2}$ (13.11)

$K_{ cm }^{\prime}=\frac{M_{X}}{M_{x}+M_{X}} K_{ lab }$ (13.12)

## Verified Solution

The available kinetic energy in the center of mass is, from Equation (13.12),

$K_{ cm }^{\prime}=\frac{12}{4+12}(14.6 MeV )=11.0 MeV$

The excitation energy $E\left({ }^{16} O ^{*}\right)$ due to just the masses when an α particle and ${ }^{12} C \text { form to make }{ }^{16} O$ is determined from Equation (13.11) to be

$E\left({ }^{16} O ^{*}\right)=M\left({ }^{4} He \right) c^{2}+M\left({ }^{12} C \right) c^{2}-M\left({ }^{16} O \right) c^{2}$

The appropriate masses are $M\left({ }^{4} He \right)=4.002603 u , M\left({ }^{12} C \right)=12.0u$, and $M\left({ }^{16} O \right)=15.994915 u$. The ground state excitation energy of ${ }^{16} O ^{*}$ becomes

\begin{aligned}E\left({ }^{16} O ^{*}\right) &=(4.002603 u +12.0 u -15.994915 u ) c^{2} \\&=0.007688 c^{2} \cdot u \\&=\left(0.007688 c^{2} \cdot u \right)\left(\frac{931.5 MeV }{c^{2} \cdot u }\right)=7.16 MeV\end{aligned}

This energy, 7.2 MeV, is indicated for $\alpha+{ }^{12} C$ in Figure 13.6. The final excitation energy in ${ }^{16} O ^{*}$ is the sum of the available $K_{ cm }^{\prime}$ and 7.2 MeV.

$E_{\text {final }}\left({ }^{16} O ^{*}\right)=11.0 MeV +7.2 MeV =18.2 MeV$