For an artificially created nuclide like , the ratio is , placing it below the zone of stability due to an excess of protons. How does this nucleus predominantly decay to achieve a stable ratio?

Question

For an artificially created nuclide like ${}_{6}^{11}C$, the $n/p$ ratio is $5/6 \approx 0.833$, placing it below the zone of stability due to an excess of protons. How does this nucleus predominantly decay to achieve a stable $n/p$ ratio?

Prashant Kotian · Accepted Answer

Correct answer: C. Nuclei lying below the stability belt have a low ratio (excess protons). To increase the ratio, a proton is converted into a neutron with the emission of a positron (): . For example, .

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