Chapter 29: Problem 63
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?
Short Answer
Expert verified
Based on the given information:
(a) The nuclear reaction for the decay of \(^{15}C\) can be written as:
$$^{15}C \rightarrow {}^{15}N + e^- + \bar{\nu_e}$$
(b) To calculate the energy released during the decay process, we first need to find the mass difference:
$$\Delta m = m_C - (m_N + m_e)$$
Then, 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 be obtained by plugging in the mass difference and the speed of light into the equation.