We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program


Advertise your business, and reach millions of students around the world.


All the data tables that you may search for.


For Arabic Users, find a teacher/tutor in your City or country in the Middle East.


Find the Source, Textbook, Solution Manual that you are looking for in 1 click.


Need Help? We got you covered.

Chapter 13

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