Question

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

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

Answer 1

Answer: (B)

$U^{235}+n\to {}_{54}^{140}Xe+{}_{38}^{94}Sr+2n$
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,
${}_{54}^{140}Xe\to {}_{55}^{140}Cs\to {}_{56}^{140}Ba\to {}_{57}^{140}La\to {}_{58}^{140}Ce$
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
${}_{38}^{94}Sr\to {}_{39}^{94}Y\to {}_{40}^{94}Zr$