Modern Physics Ques 213
- 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: 213.(a)
Solution:
Formula:
- 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, $\left|P _{xe}\right|=\left|P _{sr}\right|$
$\text { From } K=\frac{P^{2}}{2 m} \Rightarrow K \propto \frac{1}{m} $
$\quad \frac{K _{Sr}}{K _{xe}}=\frac{m _{xe}}{m _{Sr}}$
$\therefore \quad K _{Sr}=129$ $ MeV, K _{xe}=86$ $MeV$
NOTE: There is no need of finding total energy released in the process.
BETA
