A) one negative \[\beta \text{-}\]particle
B) \[\alpha \]-particle
C) a positron
D) a neutron
E) \[\gamma -\]particle
Correct Answer: D
Solution :
\[_{4}B{{e}^{9}}+{{\,}_{2}}H{{e}^{4}}\xrightarrow{{}}{{\,}_{0}}{{C}^{12}}+X\] From conservation of mass number, mass number of\[X=9+4-12=1\] Similarly, atomic number of \[X\] \[=4+2-6=0\] So, \[X\] is \[_{0}{{X}^{1}},\]i.e., neutron\[({{\,}_{0}}{{n}^{-1}})\].You need to login to perform this action.
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