Question

A fission reaction is given by $\underset{92}{236}U\to \underset{54}{140}Xe+\underset{38}{94}Sr+x+y$, where x and y are two particles. Considering $\underset{92}{236}U$ to be at rest, the kinetic energies of the products are denoted by $K_{X{e}^{\prime }},K_{S{r}^{\prime }}{K}_x\left(2MeV\right)$ and $K_y\left(2MeV\right)$, respectively. Let the binding energies per nucleon of $\underset{92}{236}U,\underset{54}{140}Xe$ and $\underset{38}{94}Sr$ be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s) is (are)

• A. $x=p,y=e^{-},K_{Sr}=129MeV,K_{Xe}=86MeV$
• B. $x=p,y=n,K_{Sr}=129MeV,K_{Xe}=86MeV$
• C. $x=n,y=n,K_{Sr}=129MeV,K_{Xe}=86MeV$
• D. $x=n,y=n,K_{Sr}=86MeV,K_{Xe}=129MeV$

Answer 1

Answer: (C)

${}_{92}^{236}U{\to }_{54}^{140}Xe{+}_{38}^{94}Sr+x+y$