The fluorine isotope \({ }_{9}^{17} \mathrm{~F}\) undergoes positron emission. (a) Write a nuclear reaction for this emission. (b) Where on the band of stability would you expect to find \({ }_{9}^{17} \mathrm{~F}\) ?

In positron emission, a proton in an unstable nucleus is converted into a neutron, and a positron is emitted. A positron is an antiparticle of the electron and has the same mass as an electron but with a positive charge. The process can be represented as: \[ p^{+} \rightarrow n + e^{+} \]

To write the nuclear reaction, we need to conserve both the number of nucleons and the charge. In the given fluorine isotope \({}_{9}^{17}\mathrm{~F}\), we have 9 protons and 17 - 9 = 8 neutrons. After the positron emission, one proton is converted into a neutron, so the resulting nucleus will have 8 protons and 9 neutrons. Since the number of protons determines the element, with 8 protons, the resulting element is oxygen (O). Therefore, we can write the nuclear reaction as: \[ {}_{9}^{17}\mathrm{F} \rightarrow {}_{8}^{17}\mathrm{O} + {}_{1}^{0}\mathrm{e^{+}} \]

The band of stability represents the region where stable nuclei are found when plotting the number of neutrons (N) against the number of protons (Z). Nuclei located above the band of stability have too many neutrons, while nuclei below the band have too few neutrons relative to protons. To determine where \({}_{9}^{17}\mathrm{~F}\) lies, we'll examine its neutron-to-proton ratio (N/Z). For \({}_{9}^{17}\mathrm{~F}\), the ratio is: \[ \frac{N}{Z} = \frac{8}{9} \approx 0.89 \] For lighter elements (Z < 20), the ideal N/Z ratio for stable nuclei is approximately 1. Since \({}_{9}^{17}\mathrm{~F}\) has an N/Z ratio of 0.89, which is less than 1, it lies below the band of stability. This means that it has too few neutrons relative to protons, making it unstable and prone to positron emission. In conclusion, a) The nuclear reaction for the positron emission of \({}_{9}^{17}\mathrm{~F}\) is: \[ {}_{9}^{17}\mathrm{F} \rightarrow {}_{8}^{17}\mathrm{O} + {}_{1}^{0}\mathrm{e^{+}} \] b) The \({}_{9}^{17}\mathrm{~F}\) isotope would be expected to be found below the band of stability due to its N/Z ratio being less than 1.