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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