(d) The given equation is $\ _{2}He^{4} + \ _{z}X^{A} \rightarrow {\ _{z + 2}Y}^{A + 3} + A$
Applying charge and mass conservation
$4 + A = A + 3 + x \Rightarrow x = 1 \Rightarrow 2 + z = z + 2 + n \Rightarrow n = 0$
Hence $A$ is a neutron
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.