The power of a nuclear power plant is $200 \, MW$. How much energy will this plant produce in one day?

The binding energy per nucleon for ${}_1^2H$ and ${}_2^4He$ are $1.1 \; MeV$ and $7.1 \; MeV$ respectively. The energy released in $MeV$ when two ${}_1^2H$ nuclei fuse to form one ${}_2^4He$ nucleus is:

$A$ nucleus of uranium decays at rest into nuclei of thorium and helium. Then:

In the nuclear fission of one nucleus of $U^{235}$, the energy released is $188 \text{ MeV}$. The energy released in the nuclear fission of $235 \text{ g}$ of $U^{235}$ is nearly (Avogadro number $= 6.02 \times 10^{23} \text{ mol}^{-1}$)

$A$ 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):

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