Q. What is the particle $x$ in the following nuclear reaction $\ _{4}^{9}Be + \ _{2}^{4}He \rightarrow \ _{6}^{12}C + x$
A) Electron
B) Proton
C) Photon
D) Neutron
Explanation
(d) $\ _{4}^{9}Be + \ _{2}^{4}He \rightarrow \ _{6}^{12}C +_{0}^{1}x$
Clearly, it is a neutron
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.