Question

A $U^{235}$ atom undergoes fission by thermal neutrons according to the following reaction
$U ^{235}+n \rightarrow{ }_{54}^{140} Xe +{ }_{38}^{94} Sr +2 n$
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)

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