Q. The nucleus $\ _{92}U^{234}$ splits exactly in half in a fission reaction in which two neutrons are released.The resultant nuclei are
A) $\ _{46}Pd^{116}$
B) $\ _{45}Rh^{117}$
C) $\ _{45}Rh^{116}$
D) $\ _{46}Pd^{117}$
Explanation
(a) The splitting of $\ _{92}U^{234}$ is as follows
$\ _{92}U^{234} \rightarrow \ _{46}X^{116}\ + \ _{46}X^{116}\ + \ 2\ _{0}n^{1}\ + \ energy$
$\therefore\ _{46}X^{116}\ \ is\ \ _{46}Pd^{116}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.