The artificially produced nuclide Carbon-11 () has an ratio of , placing it below the zone of stabil — Nuclear Chemistry and Radioactivity Chemistry Question
Question
The artificially produced nuclide Carbon-11 (${}_{6}^{11}C$) has an $n/p$ ratio of $5/6$, placing it below the zone of stability. What is its primary mode of radioactive decay?
Answer: C
💡 Solution & Explanation
Nuclei below the stability belt have a low $n/p$ ratio (excess protons). To increase the ratio, a proton converts into a neutron with the simultaneous emission of a positron (${}_{1}^{1}p \rightarrow {}_{0}^{1}n + {}_{+1}^{0}e$).
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