A) \[_{2}H{{e}^{4}}\]
B) \[_{5}{{B}^{10}}\]
C) \[_{5}{{B}^{9}}\]
D) \[_{4}B{{e}^{11}}\]
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 |
| \[_{Z}{{X}^{A}}+{{\,}_{0}}{{n}^{1}}\,\to \,{{\,}_{3}}L{{i}^{7}}+{{\,}_{2}}H{{e}^{4}}\] |
| 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, \[Z=5,\text{ }A=10\] correspond to Boron \[{{(}_{5}}{{B}^{10}})\]. |
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