A) \[_{7}{{N}^{13}}\]
B) \[_{5}{{B}^{10}}\]
C) \[_{4}B{{e}^{9}}\]
D) \[_{7}{{N}^{14}}\]
Correct Answer: D
Solution :
| Key Idea: In a nuclear reaction, mass number and atomic number remain conserved. |
| Let unknown product nucleus is \[_{Z}{{X}^{A}}\]. |
| The reaction can be written as |
| \[\underset{(Oxygen)}{\mathop{_{8}{{O}^{16}}}}\,+\underset{(deuterium)}{\mathop{_{1}{{H}^{2}}}}\,\to \,\underset{(unknown\,nucleus)}{\mathop{_{Z}{{X}^{A}}}}\,+\underset{\alpha -particle}{\mathop{_{2}H{{e}^{4}}}}\,\] |
| Conservation of mass number gives, |
| \[16+2=A+4~\,\Rightarrow A=14\] |
| Conservation of atomic number gives |
| \[8+1=Z+2\,\,\Rightarrow ~\,Z=7\] |
| Thus, the unknown product nucleus is nitrogen \[({{\,}_{7}}{{N}^{14}})\] |
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