Discussion
The Q value of \({}^{197}_{79}\mathrm{Au} + {}^{12}_{6}\mathrm{C}\) \(\to {}^{206}_{85}\mathrm{At} + ?\) in MeV is :
(Given : 1 u = 931.5 MeV/c2, M(\({}^{1}_{1}\mathrm{H}\)) = 1.007825 u, Mn = 1.008665 u, M(\({}^{12}_{6}\mathrm{C}\)) = 12.000000 u, M(\({}^{197}_{79}\mathrm{Au}\)) = 196.966543 u, M(\({}^{206}_{85}\mathrm{At}\)) = 205.986667 u)
Answer: C
The Q value is −43 MeV.
Balance the reaction first. The left side has \(A = 197 + 12 = 209\) and \(Z = 79 + 6 = 85\); astatine-206 has \(A = 206\), \(Z = 85\). The missing products carry \(A = 3\), \(Z = 0\): three neutrons. Then
\[ \begin{aligned} Q &= [M_{Au} + M_{C} - M_{At} - 3m_{n}]c^{2} \\ &= [208.966543 - 205.986667 - 3.025995]\ \text{u} \\ &= -0.046119\ \text{u} \times 931.5 \approx -43\ \text{MeV} \end{aligned} \]
The reaction is endothermic: the beam must supply at least 43 MeV (more, in the laboratory frame).
+43 MeV gets the sign backwards. −30 and 30 MeV do not come from these masses.
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