A) \[{{\,}_{5}}{{B}^{10}}\]
B) \[{{\,}_{5}}{{B}^{9}}\]
C) \[{{\,}_{5}}{{B}^{11}}\]
D) \[{{\,}_{2}}H{{e}^{4}}\]
Correct Answer: A
Solution :
\[X(n,\alpha )_{3}^{7}Li\Rightarrow {{\,}_{Z}}{{X}^{A}}+{{\,}_{0}}{{n}^{1}}\to 3L{{i}^{7}}+{{\,}_{2}}H{{e}^{4}}\] \[{{\,}_{5}}{{Z}^{10}}={{\,}_{5}}{{B}^{10}}\]
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