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 \[X=9+4-12=1.\] Similarly, atomic number of \[X=4+2-6=0\] So, X is \[{{\,}_{0}}{{X}^{1}},\]ie, neutron \[({{\,}_{0}}n{{\,}^{1}}).\]You need to login to perform this action.
You will be redirected in
3 sec