Modern Physics 5 Question 7
7. A fission reaction is given by ${ } _{92}^{236} U \rightarrow{ } _{54}^{140} Xe+{ } _{38}^{94} Sr+x+y$, where $x$ and $y$ are two particles. Considering ${ } _{92}^{236} U$ to be at rest, the kinetic energies of the products are denoted by $K _{Xe}$, $K _{Sr}, K _x(2 MeV)$ and $K _y(2 MeV)$, respectively. Let the binding energies per nucleon of ${ } _{92}^{236} U,{ } _{54}^{140} Xe$ and ${ } _{38}^{94} Sr$ be 7.5 $MeV$, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct options is/are
(2015 Adv.)
(a) $x=n, y=n, K _{Sr}=129 MeV, K _{X e}=86 MeV$
(b) $x=p, y=e^{-}, K _{Sr}=129 MeV, K _{X e}=86 MeV$
(c) $x=p, y=n, K _{Sr}=129 MeV, K _{Xe}=86 MeV$
(d) $x=n, y=n, K _{Sr}=86 MeV, K _{X _e}=129 MeV$
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Answer:
Correct Answer: 7. (c)
Solution:
- From conservation laws of mass number and atomic number, we can say that $x=n, y=n$
$$ \left(x={ } _0^{1} n, y={ } _0^{1} n\right) $$
$\therefore$ Only (a) and (d) options may be correct.
From conservation of momentum, $\left|P _{xe}\right|=\left|P _{sr}\right|$
$$ \begin{aligned} & \text { From } \quad K=\frac{P^{2}}{2 m} \Rightarrow K \propto \frac{1}{m} \\ & \frac{K _{Sr}}{K _{xe}}=\frac{m _{xe}}{m _{Sr}} \\ & \therefore \quad K _{Sr}=129 MeV, K _{xe}=86 MeV \end{aligned} $$
NOTE There is no need of finding total energy released in the process.
