A) one \[{{\beta }^{-}}\]particle
B) \[\alpha \]-particle
C) a positron
D) a neutron
Correct Answer: D
Solution :
\[_{4}B{{e}^{9}}{{+}_{2}}H{{e}^{4}}{{\xrightarrow{{}}}_{6}}{{C}^{12}}+X\] From conservation of mass number, mass number of \[X=9+4-12=L\]. Similarly, atomic number of \[X=4+2-6=0\] So, X is \[_{0}{{X}^{1}}\], ie, neutron \[{{(}_{0}}{{n}^{1}})\].
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