One nuclide in each of these pairs is radioactive. Predict which is radioactive and which is stable: \((\mathbf{a})_{20}^{40} \mathrm{Ca}\) and \(_{20}^{45} \mathrm{Ca}\) , \((\mathbf{b})^{12} \mathrm{C}\) and \(^{14} \mathrm{C},\) \((\mathbf{c})\) lead-206 and thorium-230. Explain your choice in each case.

First, let's find the number of protons (Z) and neutrons (N) in each nuclide. We can find the number of protons from the atomic number, and the number of neutrons can be determined by subtracting the atomic number from the mass number (A - Z). \[\begin{cases} (a)\\ _{20}^{40} \mathrm{Ca}: Z = 20, N = 40 - 20 = 20 \\ _{20}^{45} \mathrm{Ca}: Z = 20, N = 45 - 20 = 25 \end{cases}\] \[\begin{cases} (b)\\ ^{12} \mathrm{C}: Z = 6, N = 12 - 6 = 6 \\ ^{14} \mathrm{C}: Z = 6, N = 14 - 6 = 8 \end{cases}\] \[\begin{cases} (c)\\ \mathrm{Pb}-206: Z = 82, N = 206 - 82 = 124 \\ \mathrm{Th}-230: Z = 90, N = 230 - 90 = 140 \end{cases}\]

Now let's analyze the stability of each nuclide. (a) For \(_{20}^{40}\mathrm{Ca}\) and \(_{20}^{45}\mathrm{Ca}\), both are Isotopes of calcium and have the same number of protons (20). Since the number of neutrons and protons are equal in \(_{20}^{40}\mathrm{Ca}\), it is a stable nuclide, while \(_{20}^{45}\mathrm{Ca}\) has an excess of neutrons, making it more likely to be unstable and radioactive. (b) For \(^{12}\mathrm{C}\) and \(^{14}\mathrm{C}\), these carbon isotopes have the same number of protons (6), but different numbers of neutrons. In the case of \(^{12}\mathrm{C}\), the number of neutrons and protons are equal, which makes it a stable nuclide. \(^{14}\mathrm{C}\) has 8 neutrons, and due to the neutron excess, it is more likely to be unstable and thus radioactive. (c) For lead-206 and thorium-230, we can analyze their stability based on the ratio of neutrons to protons (N/Z). \[\frac{N}{Z}(\mathrm{Pb}-206) = \frac{124}{82} \approx 1.51\] \[\frac{N}{Z}(\mathrm{Th}-230) = \frac{140}{90} \approx 1.56\] The ideal N/Z ratio for heavy elements is around 1.5. Since the ratio for lead-206 is closer to 1.5, it is more likely to be stable, while thorium-230 is less stable and likely to be radioactive.

Based on our analysis, we can predict the radioactive nuclide in each pair: (a)\(_{20}^{40} \mathrm{Ca}\) is _stable_, and \(_{20}^{45} \mathrm{Ca}\) is _radioactive_ due to the excess of neutrons. (b)\(^{12} \mathrm{C}\) is _stable_, and \(^{14} \mathrm{C}\) is _radioactive_ due to the excess of neutrons. (c) _Lead-206_ is _stable_, and _thorium-230_ is _radioactive_ due to having an N/Z ratio farther from the ideal ratio for heavy elements.