Q. A nucleus $\ _{z}X^{A}$emits an $\alpha$-particle. The resultant nucleus emits a $\beta^{+}$particle. The respective atomic and mass numbers of the final nucleus will be
A) $Z - 3,\ A - 4$
B) $Z - 1,\ A - 4$
C) $Z - 2,\ A - 4$
D) $Z,\ A - 2$
Explanation
(b) $\ _{Z}X^{A} \longrightarrow_{\ }{\ _{Z - 2}X}^{A - 4} +_{2}{He}^{4}$
$\ _{Z - 2}X^{A - 4} \rightarrow {\ _{Z - 2}X}^{A - 4}{+ -}_{\ }1e^{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.