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?

Options

A. 50, 36
B. 58, 40
C. 56, 42
D. 57, 41

Answer

B

Explanation

`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`