The energy released per fission of a nucleus of ${}^{240}X$ is $200 \ MeV$. The energy released if all the atoms in $120 \ g$ of pure ${}^{240}X$ undergo fission is $........ \times 10^{25} \ MeV$. (Given $N_A = 6 \times 10^{23}$)

The decrease in the mass of Uranium in a nuclear reactor operating at a power of $12 \text{ MW}$ per day is (Energy released in one ${}_{92}U^{235}$ fission is about $200 \text{ MeV}$):

$U^{235}$ 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 $144 \times 10^4 \text{ s}$, then 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 $\text{ kg}$)

If the energy released per fission of a ${ }_{92}^{235} U$ nucleus is $200 \text{ MeV}$, the energy released in the fission of $0.1 \text{ kg}$ of ${ }_{92}^{235} U$ in kilowatt-hour is:

The energy released per fission of uranium-$235$ is about $200 \, MeV$. $A$ reactor using $U-235$ as fuel is producing $1000 \, kW$ power. The number of $U-235$ nuclei undergoing fission per second is approximately:

Given that the mass of ${ }_{3}^{7} Li = 7.0160 \, u$,the mass of ${ }_{2}^{4} He = 4.0026 \, u$,and the mass of ${ }_{1}^{1} H = 1.0079 \, u$. When $20 \, g$ of ${ }_{3}^{7} Li$ is converted into ${ }_{2}^{4} He$ by proton capture,the energy liberated (in $kWh$) is: [Take $1 \, u = 931.5 \, MeV/c^2$ and $1 \, kWh = 3.6 \times 10^6 \, J$]