Q. The number of neutrons released during the fission reaction is $_{0}^{1}{n +}_{92}^{235}{U \rightarrow}_{51}^{133}{Sb +_{41}^{99}{Nb +}}$neutrons
A) 1
B) 92
C) 3
D) 4
Explanation
(d) $\ _{0}n^{1} + \ _{92}U^{235} \longrightarrow \ _{51}{Sb}^{133} + \ _{41}{Nb}^{99} + Neutrons$
Charge number is conserved (92=51+41)
Applying principle of conservation of mass number
$133 + 99 + x = 235 + 1$
$x = 236 - 232 = 4$
$\therefore$ Number of neutrons $(\ _{1}n^{1})\ = 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.