A) \[_{2}^{4}He\]
B) \[_{3}^{6}Li\]
C) \[_{3}^{7}Li\]
D) \[_{4}^{8}Be\]
Correct Answer: B
Solution :
\[_{4}^{9}Be+\underset{(p)}{\mathop{_{1}{{H}^{1}}}}\,\to _{3}^{6}Li+\underset{(\alpha -particle)}{\mathop{_{2}H{{e}^{4}}}}\,\]
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