When ^{235}_{92}U is bombarded with one neutr | vedclass

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Question

(B) The nuclear fission reaction is given by the conservation of mass number and atomic number.The reaction is: $^1_0n + ^{235}_{92}U \rightarrow ^{94

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$^{139}_{56}Ba$

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(B) The nuclear fission reaction is given by the conservation of mass number and atomic number.The reaction is: $^1_0n + ^{235}_{92}U \rightarrow ^{94}_{36}Kr + X + 3(^1_0n)$.Let the unknown product $X$ be $^{A}_{Z}Y$.Equating the mass numbers: $1 + 235 = 94 + A + 3(1) \Rightarrow 236 = 97 + A \Rightarrow A = 139$.Equating the atomic numbers: $0 + 92 = 36 + Z + 3(0) \Rightarrow 92 = 36 + Z \Rightarrow Z = 56$.The element with atomic number $56$ is Barium $(Ba)$.Thus,the product is $^{139}_{56}Ba$.

When $^{235}_{92}U$ is bombarded with one neutron,fission occurs and the products are three neutrons,$^{94}_{36}Kr$ and:

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