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