Question

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 ${ }_{92}^{236} U$ being at rest, the kinetic energies of the products are denoted by ${K}_{X e}, K_{S r}, K_x(2 \mathrm{~MeV})$ and $K_y(2 \mathrm{~MeV}),$ respectively. Let the binding energies per nucleon of ${ }_{92}^{236} {U},{ }_{54}^{140} {Xe},{ }_{38}^{94} {Sr}$ be $7.5 \mathrm{~MeV}, 8.5 \mathrm{~MeV}$ and $8.5 \mathrm{~MeV}$ respectively. Considering different conservation laws, the correct option is

Accepted Answer

$x=n, y=n, K_{S r}=129 \mathrm{~MeV}, K_{X e}=86 \mathrm{~MeV}$

Answer

$x=p, y=e^{-}, K_{S r}=129 \mathrm{~MeV}, K_{X e}=86 \mathrm{~MeV}$

Answer

$x=p, y=n, K_{S r}=129 \mathrm{~MeV}, K_{X e}=86 \mathrm{~MeV}$

Answer

$x=n, y=n, K_{S r}=86 \mathrm{~MeV}, K_{X e}=129 \mathrm{~MeV}$

NEET UG PHYSICS MCQ: NUCLEI

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