When a slow neutron goes sufficiently close to a ${U^{235}}$ nucleus,then the process that takes place is

Fission of nuclei is possible because the binding energy per nucleon in them

$A$ nuclear reactor with $40\%$ efficiency produces $10^{14}$ fissions per second. Each fission releases $250 \text{ MeV}$ of energy. The power output of the reactor is ......... $W$.

When a proton collides with a $_3Li^7$ nucleus and results in a $_4Be^8$ nucleus,the emitted particle is:

If $200 \text{ MeV}$ of energy is released in the fission of one nucleus of ${ }_{92}^{236} U$, the number of nuclei that must undergo fission to release an energy of $1000 \text{ J}$ is

The average energy released per fission for the nucleus of $_{92}^{235} U$ is $190 \text{ MeV}$. When all the atoms of $47 \text{ g}$ pure $_{92}^{235} U$ undergo fission process, the energy released is $\alpha \times 10^{23} \text{ MeV}$. The value of $\alpha$ is . . . . . . . . . . . (Avogadro Number $= 6 \times 10^{23} \text{ per mole}$)