The number of neutrons in the parent nucleus | vedclass

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(D) The $\beta$-emission process is represented as: $_{Z}^{A}X \rightarrow _{Z+1}^{A}Y + _{-1}^{0}e$. Given that the daughter nucleus is $^{14}N$ $(_{

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$8$

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(D) The $\beta$-emission process is represented as: $_{Z}^{A}X \rightarrow _{Z+1}^{A}Y + _{-1}^{0}e$. Given that the daughter nucleus is $^{14}N$ $(_{7}^{14}N)$,the parent nucleus must have an atomic number $Z = 7 - 1 = 6$ and mass number $A = 14$. Thus,the parent nucleus is $^{14}C$ $(_{6}^{14}C)$. The number of neutrons in the parent nucleus is calculated as $A - Z = 14 - 6 = 8$.

The number of neutrons in the parent nucleus which gives $^{14}N$ on $\beta$-emission is

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