Answer
a. Carbon-12.
b. 5.702 MeV.
Work Step by Step
a. On the left side of the reaction, the total number of nucleons is 13. This must also be true on the right hand side. The neutron is 1 nucleon, so the mystery nucleus has 12 nucleons.
On the left side of the reaction, the total number of protons is 2+4=6. This must also be true on the right hand side. The neutron is neutral, so the mystery nucleus has 6 protons.
The mystery nucleus must be carbon-12.
b. Calculate the Q-value, using data from Appendix B. We use atomic masses.
$$Q=\left(m_{\alpha}+m(^{9}_{4}Be)-m(^{12}_{6}C)-m_n \right)c^2$$
$$ =\left(4.002603u+9.012183u-12.000000u-1.008665u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$=5.702MeV$$
The Q-value is positive. The reaction is exothermic, and releases 5.702 MeV of energy.
