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 $= 6 \times 10^{23} \text{ per mole}$)

$A$ $^{235}U$ nuclear reactor generates energy at a rate of $3.70 \times 10^7 \text{ J/s}$. Each fission liberates $185 \text{ MeV}$ of useful energy. If the reactor has to operate for $144 \times 10^4 \text{ s}$, the mass of the fuel needed is (Assume Avogadro's number $= 6 \times 10^{23} \text{ mol}^{-1}$, $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$) (in $\text{ kg}$)

On fission,a $U^{235}$ nucleus releases $3 \times 10^{-11} \, J$ of energy. In a $1 \, GW$ nuclear reactor,$4.2 \%$ of this energy is converted to useful energy. The $U^{235}$ consumed (in grams) in half an hour is closest to (Avogadro number $N_A = 6.023 \times 10^{23}$)

$A$ nucleus of mass $M + \Delta m$ is at rest and decays into two daughter nuclei of mass $\frac{M}{4}$ and $\frac{3M}{4}$. If the speed of light is $c$,the speed of the daughter nucleus of mass $\frac{M}{4}$ is:

The phenomenon in which a proton flips is

How can fusion reactors provide unlimited power to humanity?