If in a nuclear fission,a piece of uranium of mass $0.5\, g$ is lost,the energy obtained in $kWh$ is

In a nuclear reactor, the main purpose of the moderator is to

Thermal neutrons are those which

In a fission reaction $_{92}^{236}U \to ^{117}X + ^{117}Y + n + n$,the binding energy per nucleon of $X$ and $Y$ is $8.5\, MeV$,whereas that of $^{236}U$ is $7.6\, MeV$. The total energy liberated will be about . . . . . . $MeV$.

Nuclear fission was discovered by

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 $1 \, kWh = 3.6 \times 10^6 \, J$]