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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