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## Q. 13.3

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).

## Verified Solution

The ground state Q value is calculated from Equation (13.7).

\begin{aligned}\frac{Q}{c^{2}}=& M\left({ }^{4} He \right)+M\left({ }^{14} N \right)-\left[M\left({ }^{1} H \right)+M\left({ }^{17} O \right)\right] \\=& 4.002603 u +14.003074 u \\&-(1.007825 u +16.999132 u ) \\=&-0.001280 u\end{aligned}

$Q=\left(-0.001280 c^{2} \cdot u \right)\left(\frac{931.5 MeV }{c^{2} \cdot u }\right)=-1.192 MeV$

\begin{aligned}Q &=K(p)+K\left({ }^{17} O \right)-K(\alpha) \\K(p)+K\left({ }^{17} O \right) &=Q+K(\alpha) \\&=-1.192 MeV +7.7 MeV =6.5 MeV\end{aligned}

The final reaction products share 6.5 MeV of energy.