The artificially produced isotope Carbon-11 () has an ratio of (), placing it below the belt of stab — Nuclear Chemistry and Radioactivity Chemistry Question
Question
The artificially produced isotope Carbon-11 (${}_{6}^{11}C$) has an $n/p$ ratio of $5/6$ ($\approx 0.83$), placing it below the belt of stability. By which primary decay mechanism will this nuclide attempt to reach the stable zone?
Answer: D
💡 Solution & Explanation
Nuclei lying below the stability belt have too few neutrons (low $n/p$). To increase the ratio, a proton converts into a neutron with the emission of a positron: ${}_{1}^{1}p \rightarrow {}_{0}^{1}n + {}_{+1}^{0}e$.
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