Question

Consider one of fission reactions of ${}^{235}U$ by thermal neutrons ${}_{92}^{235}U+n\to {}_{38}^{94}Sr+{}_{54}^{140}Xe+2n$. 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: $m\left({}^{235}U\right)=235.439u,m(n)=1.00866u,m\left({}^{94}Zr\right)=93.9064u,m\left({}^{140}Ce\right)=139.9055u,1u=931MeV]$

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

Answer 1

Answer: (B)

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