The missing particle in the reaction,_{92}^{2 | vedclass

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(C) In a nuclear fission reaction,both the mass number and the atomic number must be conserved on both sides of the equation.Let the missing particle

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$_{36}^{87}Kr$

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(C) In a nuclear fission reaction,both the mass number and the atomic number must be conserved on both sides of the equation.Let the missing particle be $_{Z}^{A}X$.The equation is: $_{92}^{235}U + _{0}^{1}n 	o _{56}^{146}Ba + _{Z}^{A}X + 3_{0}^{1}n$.Conservation of mass number: $235 + 1 = 146 + A + 3(1) \implies 236 = 149 + A \implies A = 87$.Conservation of atomic number: $92 + 0 = 56 + Z + 3(0) \implies 92 = 56 + Z \implies Z = 36$.The element with atomic number $36$ is Krypton $(Kr)$.Thus,the missing particle is $_{36}^{87}Kr$.

The missing particle in the reaction,$_{92}^{235}U + _{0}^{1}n \to _{56}^{146}Ba + ... + 3_{0}^{1}n$ is

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