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

Prashant Kotian · Accepted Answer

Correct answer: 9. The reaction path is Parent . Each emission increases the atomic number by 1 while keeping mass number constant. Thus, the parent must have an atomic number . The mass number of the parent remains . The parent nucleus is . The number of neutrons in this parent nucleus is .

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