Question

${}_6^{14}C{\to }_7^{14}N+X$
What is X?

• A. ${}_2^{4}He$
• B. ${}_{+1}^{0}e$
• C. ${}_{-1}^{0}e$
• D. ${}_1^{1}H$
• E. ${}_6^{12}C$

Answer 1

Answer: (C)

${}_{-1}^{0}e$

$X$ must be ${}_{-1}^{0}e$, because we can easily see from the radioactive reaction that the mass number of the two elements formed before and after the reaction remain the same, that is $14$. Whereas, the number of protons of the element on the right hand side has increased by $1$. Thus, the particle which is produced along with nitrogen, must have a mass number of $0$ and its number of protons must be $-1$ to balance the the number of protons on both the sides of the reaction.

Option C is the correct answer.