A U^(235) atom undergoes fission by thermal neutrons according to the following reaction U^(235)+nrarr""_(54)^(140)Xe+""_(38)^(94)Sr+2n Then Xenon undergoes four and Strontium undergoes two consecutive B decays and six electrons are detected. What is the atomic number of the two decay products of Xenon and Strontium?

A U^(235) atom undergoes fission by thermal neutrons according to the

1.	A U^(235) atom undergoes fission by thermal neutrons according to the following reaction U^(235)+nrarr""_(54)^(140)Xe+""_(38)^(94)Sr+2n Then Xenon undergoes four and Strontium undergoes two consecutive B decays and six electrons are detected. What is the atomic number of the two decay products of Xenon and Strontium?
Answer» 50, 36 58, 40 56, 42 57, 41 Solution :`U^(235)+n rarr ""_(54)^(140)Xe+""_(38)^(94)Sr+2n` When `beta` particle is emitted, atomic number of daughter NUCLEI. is increased by 1 and atomic MASS number remains the same. Therefore, when `""_(54)^(140)Xe `undergoes four consecutive `beta` decays, the atomic number of final decay product is increased by 4 i.e., 58 and corresponding decay CHAIN is shown here, `""_(54)^(140)Xerarr""_(55)^(140)CS rarr ""_(56)^(140)Bararr""_(57)^(140)La rarr ""_(58)^(140)Ce` When `""_(38)^(94)Sr` undergoes two consecutive `beta` decays, the atomic number of final decay product is increased by 2. i.e. 40 and corresponding decay chain is shown below `""_(38)^(94)Sr rarr ""_(39)^(94)Y rarr ""_(40)^(94)Zr`

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