For an artificially created nuclide like , the ratio is , placing it below the zone of stability due — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Nuclei lying below the stability belt have a low $n/p$ ratio (excess protons). To increase the ratio, a proton is converted into a neutron with the emission of a positron ($\beta^+$): ${}_{1}^{1}p \rightarrow {}_{0}^{1}n + {}_{+1}^{0}e$. For example, ${}_{6}^{11}C \rightarrow {}_{5}^{11}B + {}_{+1}^{0}e$.