A) \[_{13}^{27}Al+_{2}^{4}He\to _{15}^{30}P\]
B) \[_{6}^{27}C+_{1}^{1}H\to _{7}^{13}N\]
C) \[_{15}^{30}P\to _{14}^{30}Si\]
D) \[_{96}^{241}Am+_{2}^{4}He\to _{97}^{245}Bk\]
Correct Answer: A
Solution :
\[_{13}^{27}Al+_{2}^{4}He\xrightarrow{{}}_{15}^{30}P+_{0}^{1}n\]
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