A) \[_{0}{{n}^{1}}\] emission
B) \[\beta -\]emission
C) \[K-\]electron capture
D) \[\alpha -\]emission
Correct Answer: A
Solution :
\[_{11}{{N}^{24}}{{\xrightarrow{{}}}_{11}}{{N}^{23}}{{+}_{0}}{{n}^{1}}\] Because of emission of neutron, \[_{11}Na\]is formed.
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