Question

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

• A. $156MeV$
• B. $208MeV$
• C. $456MeV$
• D. Cannot be computed

Answer 1

The correct option is B $208MeV$

Energy released $=△M{C}^2=△M\times 931MeV$

$△M=M_{components}-M_{products}$

$=(235.439+1.00866)-(93.9064+139.9055+2(1.00866\left)\right)$

$△M=0.22$

$\therefore E=0.22\times 931=204.82\simeq 208MeV$