The artificially produced isotope Carbon-11 () has an ratio of (), placing it below the belt of stability. By which primary decay mechanism will this nuclide attempt to reach the stable zone?

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?

Prashant Kotian · Accepted Answer

Correct answer: D. Nuclei lying below the stability belt have too few neutrons (low ). To increase the ratio, a proton converts into a neutron with the emission of a positron: .

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