A) \[\beta \]-active, \[_{7}{{N}^{14}}\]
B) \[\alpha \]- active, \[_{7}B{{e}^{10}}\]
C) Positron active, \[_{5}{{B}^{14}}\]
D) \[\gamma \]- active, \[{{C}^{14}}\]
Correct Answer: A
Solution :
\[_{6}{{C}^{14}}\to {{\,}_{7}}{{N}^{14}}{{+}_{-1}}{{e}^{o}}\], b-active.
You need to login to perform this action.
You will be redirected in
3 sec
