In a nuclear fission reaction given below, the three neutrons are released. The value of $x$ and $y$ are
${^1_0n}+{^{235}_{92}U} → {^x_{56}Ba} + {^{89}_yKr}+ 3{^1_0n}$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $x = 144, y = 36$
Given: $^{1}_{0}\!n + \,^{235}_{92}\!U \;\to\; ^{x}_{56}\!Ba \;+\; ^{89}_{y}\!Kr \;+\; 3\,^{1}_{0}\!n$
Mass-number conservation: $1+235 = x + 89 + 3\cdot 1 \;\Rightarrow\; 236 = x + 92 \;\Rightarrow\; x = 144$
Atomic-number conservation: $0+92 = 56 + y + 3\cdot 0 \;\Rightarrow\; 92 = 56 + y \;\Rightarrow\; y = 36$
Result: $x=144,\; y=36$