The carbon isotope \({ }^{15} \mathrm{C}\) decays much faster than ${ }^{14} \mathrm{C}$. (a) Using Appendix B, write a nuclear reaction showing the decay of ${ }^{15} \mathrm{C}\(. (b) How much energy is released when \){ }^{15} \mathrm{C}$ decays?

The first step is to look at Appendix B mentioned in the question to identify the decay mode of \(^{15}C\). Appendix B should contain the decay scheme of isotopes. From this appendix, we can find that \(^{15}C\) decays via beta emission, where a neutron is converted into a proton and a beta particle (electron) is emitted. The nuclear reaction for the decay of \(^{15}C\) can be written as: $$^{15}C \rightarrow {}^{15}N + e^- + \bar{\nu_e}$$ Here, \(^{15}C\) decays into Nitrogen (\(^{15}N\)), an electron (beta particle, \(e^-\)) and an anti-neutrino (\(\bar{\nu_e}\)).

To calculate the energy released during the decay process, we need the masses of all particles involved. The masses should also be available in the given appendix. For this calculation purpose, let's assume the masses of the particles involved are the following: - Mass of \(^{15}C (m_C)\) - Mass of \(^{15}N (m_N)\) - Mass of electron (beta particle) \(m_e\) We can ignore the mass of the anti-neutrino since it has negligible mass compared to other particles involved. The conservation of mass-energy states that total mass-energy before decay should be equal to the total mass-energy after decay. To find the energy released, we can use the mass-energy equivalence formula, E=mc^2, where E is the released energy, m is the mass difference between input and output particles, and c is the speed of light. First, we need to find the mass difference: $$\Delta m = m_C - (m_N + m_e)$$ Next, we can calculate the energy released using E=mc^2: $$E = (\Delta m)c^2$$ The energy released during the decay of \(^{15}C\) can then be obtained by plugging in the mass difference and the speed of light into the equation.