A highly unstable parent nucleus undergoes two successive emissions to directly form the completely stable isotope . What was the original number of neutrons present in the initial parent nucleus?

Question

A highly unstable parent nucleus undergoes two successive $\beta^-$ emissions to directly form the completely stable isotope ${}_{7}^{14}N$. What was the original number of neutrons present in the initial parent nucleus?

Prashant Kotian · Accepted Answer

Correct answer: C. The reaction path is Parent . Each emission increases by 1 while keeping mass number constant. Reversing the reaction drops by 2. Thus, parent and . The parent is . Neutrons = .

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