1. Tardigrade
  2. Question
  3. Physics
  4. Consider one of fission reactions of 235 U by thermal neutrons 92235 U +n arrow 3894 Sr + 54140 Xe +2 n. The fission fragments are however unstable and they undergo successive β-decay until 3894 Sr becomes 4094 Zr and 54140 Xe becomes 58140 Ce. The energy released in this process is [Given: m( 235 U )=235.439 u , m( n )=1.00866 u , m( 94 Zr ).=93.9064 u , m( 140 Ce )=139.9055 u , 1 u =931 MeV ]

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Consider one of fission reactions of ${ }^{235} U$ by thermal neutrons ${ }_{92}^{235} U +n \rightarrow{ }_{38}^{94} Sr +{ }_{54}^{140} Xe +2 n$. The fission fragments are however unstable and they undergo successive $\beta$-decay until ${ }_{38}^{94} Sr$ becomes ${ }_{40}^{94} Zr$ and ${ }_{54}^{140} Xe$ becomes ${ }_{58}^{140} Ce$. The energy released in this process is
[Given: $m\left({ }^{235} U \right)=235.439\, u , m( n )=1.00866\, u , m\left({ }^{94} Zr \right)\left.=93.9064\, u , m\left({ }^{140} Ce \right)=139.9055\, u , 1\, u =931\, MeV \right]$

Nuclei

Solution:

The complete fission reaction is
${ }_{92}^{235} U +n \rightarrow{ }_{40}^{94} Zr +{ }_{58}^{140} Ce +2 n+6 e^{-1}$
$Q=\left[m\left({ }^{235} U \right)-m\left({ }^{94} Zr \right)-m\left({ }^{140} Ce \right)-m(n)\right] c^{2}$
$=208\, MeV$