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

Answer 1

Answer: (A)

$x=n,y=n,K_{Sr}=129MeV,K_{Xe}=86MeV$

${}_{92}^{236}U{\to }_{54}^{140}Xe{+}_{38}^{94}Sr+x+y$
From conservation of charge, the combined charges of x and y should be zero since LHS has a charge of 92 and RHS has already charge of 92.
Hence, option(C) is ruled out since combined charge of p and n is not zero.
Change in $BE=-236\times 7.5+140\times 8.5+94\times 8.5=219Mev$
This difference is equal to combined KE of the products from fission.
Momentum is conserved and is zero since initial momentum is zero.
Case A:
Both x and y have the same KE. They will have same momentum since masses are same.
Sr is lighter, hence should have more KE.
This is satisfied.
Case D:
Sr has less KE, it is not possible.
Case B: Lepton no conservation will not allow B( electron is a lepton).