Complete the reaction: $n + _{92}^{235}U \to _{56}^{144}Ba + .... + 3n$

The thermonuclear reaction of hydrogen inside the stars takes place by a cycle of operations. The particular element which acts as a catalyst is

$A$ 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 option$(s)$ is(are):

Energy in the sun is generated mainly by

Nuclear fission is best explained by
[$AIPMT$ $2000$]

When neutrons are bombarded on the nucleus of $_{92}^{235}U$,the number of emitted neutrons is typically