A) 2He4
B) 5B10
C) 5B9
D) 4Be11
Correct Answer: B
Solution :
Key Idea: In a nuclear reaction conservation of both the charge number and mass number must hold. The given nuclear reaction can be written as\[_{Z}{{X}^{A}}{{+}_{0}}{{n}^{1}}{{\xrightarrow[{}]{{}}}_{3}}L{{i}^{7}}{{+}_{2}}H{{e}^{4}}\] Conservation of mass number gives, \[A+1=7+4\] \[\Rightarrow \] \[A=10\] Conservation of charge number gives, \[Z+0=2+3\] \[\Rightarrow \] \[Z=5\] Hence, \[Z=5,\,A=10\]correspond to Boron \[{{(}_{5}}{{B}^{10}}).\]
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