In the neutron-induced fission reaction $^{1}_{0} ext{n} + ^{235}_{92} ext{U} ightarrow ^{236}_{

Question

In the neutron-induced fission reaction $^{1}_{0}	ext{n} + ^{235}_{92}	ext{U} ightarrow ^{236}_{92}	ext{U} ightarrow ^{144}_{56}	ext{Ba} + ^{89}_{36}	ext{Kr} + x ^{1}_{0}	ext{n}$, the value of $x$ (number of neutrons produced) is:

Professor PC Thomas Classes · Answer

By balancing the mass numbers: $1 + 235 = 144 + 89 + x(1)$. Thus, $236 = 233 + x \implies x = 3$.