Question

${}_6^{14}C{\to }_7^{14}N+X$
In the above reaction, 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)

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

Given ${}_6^{14}C{⟶}_7^{14}N+X$

Atomic mass and mass balance gives

${}_6^{14}C{⟶}_7^{14}{N+}_{-1}^{0}X$

The $X$ is Electron ${}_{-1}^{0}e$