A fission reaction is given by where and are two particles. Considering to be at rest, the kinetic energies of the products are denoted by and , respectively. Let the binding energies per nucleon of , and be , and , respectively. Considering diffe

Question

A fission reaction is given by ${}_{92}^{236}U \to {}_{54}^{140}{Xe} + {}_{38}^{94}{Sr} + x + y,$ 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 option(s) is(are) [2015]

Smart Achievers · Accepted Answer

(1)

For the given fission reaction,

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

From conservation laws of mass number (A) and atomic number (Z),

${}_{92}^{236}U$ $y = n$ i.e., $x = {}_{0}^{1}n$ and $y = {}_{0}^{1}n$

From conservation of momentum, $| P_{Xe} |$ $=$ $| P_{Sr} |$

From $K = \frac{P^{2}}{2 m} \Rightarrow K \propto \frac{1}{M}$

$∴$ $\frac{K_{Sr}}{K_{Xe}} = \frac{m_{Xe}}{m_{Sr}}$

$∴$ $K_{Sr} = 129 MeV$ and $K_{Xe} = 86 MeV$

Solution

1 $x = n, y = n, K_{Sr} = 129 MeV, K_{Xe} = 86 MeV$

2 $x = p, y = e^{-}, K_{Sr} = 129 MeV, K_{Xe} = 86 MeV$

3 $x = p, y = n, K_{Sr} = 129 MeV, K_{Xe} = 86 MeV$

4 $x = n, y = n, K_{Sr} = 86 MeV, K_{Xe} = 129 MeV$

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Solution

1 x=n, y=n, KSr=129 MeV, KXe=86 MeV

2 x=p, y=e-, KSr=129 MeV, KXe=86 MeV

3 x=p, y=n, KSr=129 MeV, KXe=86 MeV

4 x=n, y=n, KSr=86 MeV, KXe=129 MeV

Ans. (1) For the given fission reaction, U92236⟶Xe54140+Sr3894+x+y From conservation laws of mass number (A) and atomic number (Z), x=n, y=n i.e., x=n01 and y=n01 From conservation of momentum, |PXe|=|PSr| From K=P22m⇒K∝1M ∴ KSrKXe=mXemSr ∴ KSr=129 MeV and KXe=86 MeV

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1 $x = n, y = n, K_{Sr} = 129 MeV, K_{Xe} = 86 MeV$

2 $x = p, y = e^{-}, K_{Sr} = 129 MeV, K_{Xe} = 86 MeV$

3 $x = p, y = n, K_{Sr} = 129 MeV, K_{Xe} = 86 MeV$

4 $x = n, y = n, K_{Sr} = 86 MeV, K_{Xe} = 129 MeV$