When a uranium isotope { }_{92}^{235} U is bo | vedclass

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(B) The nuclear fission reaction is given by the conservation of mass number and atomic number.Let the unknown nucleus be ${ }_{Z}^{A} X$.The reaction

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${ }_{56}^{144} Ba$

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(B) The nuclear fission reaction is given by the conservation of mass number and atomic number.Let the unknown nucleus be ${ }_{Z}^{A} X$.The reaction is: ${ }_{92}^{235} U + { }_{0}^{1} n \rightarrow { }_{36}^{89} Kr + { }_{Z}^{A} X + 3({ }_{0}^{1} n)$.Conservation of mass number $(A)$: $235 + 1 = 89 + A + 3(1) \Rightarrow 236 = 92 + A \Rightarrow A = 144$.Conservation of atomic number $(Z)$: $92 + 0 = 36 + Z + 3(0) \Rightarrow 92 = 36 + Z \Rightarrow Z = 56$.The element with atomic number $56$ is Barium $(Ba)$.Thus,the missing nucleus is ${ }_{56}^{144} Ba$.

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

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