Q. In the given nuclear reaction $A,B,C,D,E$ represents $\ _{92}U^{238}{\overset{\ \ \alpha\ \ }{\rightarrow}}_{B}T_{\ }h^{A}{\overset{\ \ \beta\ \ }{\rightarrow}}_{D}Pa^{C}\overset{\ \ \ e\ \ \ }{\rightarrow}\ _{92}U^{234}$
A) $A = 234,B = 90,C = 234,D = 91,E = \beta$
B) $A = 234,B = 90,C = 238,D = 94,E = \alpha$
C) $A = 238,B = 93,C = 234,D = 91,E = \beta$
D) $A = 234,B = 90,C = 234,D = 93,E = \alpha$
Explanation
(a) $\ _{92}U^{238}\overset{\ \ \alpha\ \ }{\rightarrow}\ _{90}T_{\ }h^{234}\overset{\ \ \beta\ \ }{\rightarrow}\ _{91}Pa^{234}\overset{\ \ e\left( \ _{- 1}\beta^{0} \right)\ }{\rightarrow}\ _{92}U^{234}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.