An artificial radioactive isotope gave _7^{14 | vedclass

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(A) Let the parent nucleus be $_Z^A X$.Two successive $\beta$-particle emissions are represented as: $_Z^A X \xrightarrow{2 \beta} {}_7^{14} N$.Each $

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) Let the parent nucleus be $_Z^A X$.Two successive $\beta$-particle emissions are represented as: $_Z^A X \xrightarrow{2 \beta} {}_7^{14} N$.Each $\beta$-emission increases the atomic number by $1$ while the mass number remains constant.Therefore,$Z + 2 = 7$,which gives $Z = 5$.The mass number $A$ remains $14$.The parent nucleus is $_5^{14} B$.The number of neutrons $N = A - Z = 14 - 5 = 9$.

An artificial radioactive isotope gave $_7^{14}N$ after two successive $\beta$-particle emissions. The number of neutrons in the parent nucleus must be

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