A) \[{}_{7}{{N}^{14}}\]
B) \[{}_{7}{{N}^{13}}\]
C) \[{}_{5}{{B}^{13}}\]
D) \[{}_{6}{{C}^{13}}\]
E) \[{}_{5}{{B}^{12}}\]
Correct Answer: B
Solution :
\[\underset{(Neutron)}{\mathop{{}_{6}{{C}^{12}}+{}_{0}{{n}^{1}}}}\,\xrightarrow{{}}{}_{7}{{N}^{13}}+\underset{\begin{smallmatrix} (Beta \\ particle) \end{smallmatrix}}{\mathop{{}_{-1}{{e}^{0}}}}\,+\underset{\begin{smallmatrix} (Aanti\text{-} \\ neutrino) \end{smallmatrix}}{\mathop{\overline{v}}}\,\] On equating atomic numbers and atomic masses, the atomic number and atomic mass for resulting nucleus is 7 and 13, which is for nitrogen nucleus.
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