Problem 6.265

Find the energy of the reaction $\mathrm{N}^{14}(\alpha, p) \mathrm{O}^{17}$ if the kinetic energy of the incoming alpha-particle is $T_{\alpha}=4.0 \mathrm{MeV}$ and the proton outgoing at an angle $\theta=60^{\circ}$ to the motion direction of the alpha-particle has a kinetic energy $T_{p}=2.09 \mathrm{MeV}$ .

Reveal Answer

\begin{array}{l} \text { 6.265. } Q=\left(1+\eta_{p}\right) T_{p}-\left(1-\eta_{\alpha}\right) T_{\alpha}-2 \sqrt{\eta_{p} \eta_{\alpha} T_{p} T_{\alpha}} \times \\ \times \cos \theta=-1.2 \mathrm{MeV}, \text { where } \eta_{p}=m_{\mathrm{p}} / m_{\mathrm{O}, \eta_{\alpha}=m_{\alpha}} / m_{\mathrm{o}} \end{array}

Nuclear Reactions