A) Electron
B) Positron
C) Proton
D) Neutron
Correct Answer: D
Solution :
The given equation is \[_{2}H{{e}^{4}}{{+}_{z}}{{X}^{A}}{{\to }_{z+2}}{{Y}^{A+3}}+A\] Applying charge and mass conservation \[4+A=A+3+x\Rightarrow x=1\]Þ \[2+z=z+2+n\Rightarrow n=0\] Hence A is a neutron.You need to login to perform this action.
You will be redirected in
3 sec