The fission properties of $_{94}^{239} Pu$ are very similar to those of $_{92}^{235} U$. The average energy released per fission is $180 \; MeV$. How much energy, in $MeV$, is released if all the atoms in $1 \; kg$ of pure $_{94}^{239} Pu$ undergo fission?

The fission of a $_{92}U^{235}$ nucleus releases $200 \, MeV$ of energy. Find the rate of fission of $_{92}U^{235}$ required to operate a reactor at a constant power of $5 \, W$.

$A$ $^{235}U$ 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 $\text{ kg}$)

Energy is released in nuclear fission due to:

$A$ star initially has $10^{40}$ deuterons. It produces energy via the processes:
$_1H^2 + _1H^2 \to _1H^3 + p$
$_1H^2 + _1H^3 \to _2He^4 + n$
The masses of the nuclei are as follows:
$M(H^2) = 2.014 \, amu; \, M(p) = 1.007 \, amu;$
$M(n) = 1.008 \, amu; \, M(He^4) = 4.001 \, amu$
If the average power radiated by the star is $10^{16} \, W$, the deuteron supply of the star is exhausted in a time of the order of:

The phenomenon in which a proton flips is