When a uranium isotope $^{235}_{92} ext{U}$ is bombarded with a neutron, it generates $^{89}_{36} ext{Kr}$,

Question

When a uranium isotope $^{235}_{92}	ext{U}$ is bombarded with a neutron, it generates $^{89}_{36}	ext{Kr}$, three neutrons and :

Accepted Answer

The reaction is $^{235}_{92}	ext{U} + ^{1}_{0}	ext{n} ightarrow ^{89}_{36}	ext{Kr} + 3^{1}_{0}	ext{n} + ^{A}_{Z}	ext{X}$. Balancing mass number: $235 + 1 = 89 + 3 + A \Rightarrow A = 144$. Balancing atomic number: $92 + 0 = 36 + Z \Rightarrow Z = 56$. Thus, $^{144}_{56}	ext{Ba}$ is generated.

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