Calculate the ground state Q value for the reaction { }^{14} N (\alpha, p){ }^{17} O in which Rutherford first observed a nuclear reaction. The kinetic energy of the a particles was 7.7 MeV. What was the sum of the kinetic energies of the exit channel?
Strategy We can determine the Q value with Equation (13.7) by using the atomic masses listed in Appendix 8. They are
Q=M_{x} c^{2}+M_{X} c^{2}-\left(M_{y} c^{2}+M_{Y} c^{2}\right)=K_{y}+K_{Y}-K_{x} (13.7)
\begin{array}{rlrl}M\left({ }^{4} He \right) & =4.002603 u & M\left({ }^{1} H \right) & =1.007825 u \\M\left({ }^{14} N \right) & =14.003074 u & M\left({ }^{17} O \right) & =16.999132 u\end{array}Notice that we must use the atomic masses for { }^{1} H \text { and }{ }^{4} He in order to have the complete cancellation of the electron masses. We can also determine the kinetic energies of the exit channel from Equation (13.7).