Nuclear fission and fusion can be explained on the basis of

In a reactor, $2 \, kg$ of ${ }_{92} U ^{235}$ fuel is fully used up in $30$ days. The energy released per fission is $200 \, MeV$. Given that the Avogadro number, $N_A = 6.023 \times 10^{26} \, \text{per kilo mole}$ and $1 \, eV = 1.6 \times 10^{-19} \, J$. The power output of the reactor is close to $..... \, MW$.

$A$ nuclear reactor delivers a power of $10^9 \,W$. The amount of fuel consumed by the reactor in one hour is: (in $\,g$)

The binding energy of deuteron $_1^2H$ is $1.112 \, MeV$ per nucleon and an $\alpha$-particle $_2^4He$ has a binding energy of $7.047 \, MeV$ per nucleon. Then in the fusion reaction $_1^2H + _1^2H \to _2^4He + Q$,the energy $Q$ released is ........ $MeV$.

The binding energies per nucleon for a deuteron $(_{1}H^{2})$ and an $\alpha$-particle $(_{2}He^{4})$ are $x_1$ and $x_2$ respectively. What will be the energy $Q$ released in the reaction $_{1}H^{2} + _{1}H^{2} \to _{2}He^{4} + Q$?

If the average number of neutrons liberated per fission is $2.5$ and energy released per fission is $250 \, MeV$, then the number of neutrons generated per second in a nuclear reactor of $100 \, MW$ will be: