Q. $_{92}^{238}U$ has 92 protons and 238 nucleons. It decays by emitting an alpha particle and becomes
A) $_{92}^{234}U$
B) $_{90}^{234}{Th}$
C) $_{92}^{235}U$
D) $_{93}^{237}{Np}$
Explanation
(b) Let the daughter nucleus be$\ _{Z}^{A}X$ . So, reaction can be shown as
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{92}^{238}U \rightarrow_{Z}^{A}X +_{2}^{4}{He}$
From conservation of atomic mass
$238 = \ A + 4$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ A = 234$
From conservation of atomic number
$\ \ \ \ \ \ \ 92 = Z + 2$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ Z = 90$
So, the resultant nucleus $is\ \ _{90}^{234}{\ X}\ ,\ ie,\ _{90}^{234}{\ Th}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.