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:

Slowing down of neutrons: In a nuclear reactor,a neutron of high speed (typically $10^{7} \; m s^{-1}$) must be slowed to $10^{3} \; m s^{-1}$ so that it can have a high probability of interacting with isotope $^{235}_{92}U$ and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nucleus like deuterium or carbon,which has a mass of only a few times the neutron mass. The material making up the light nuclei,usually heavy water $(D_{2}O)$ or graphite,is called a moderator.

How much $MeV$ energy is emitted by the fission of one nucleus of uranium-$235$ (in $MeV$)?

When neutrons are bombarded on the nucleus of $_{92}^{235}U$,the number of emitted neutrons is typically

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

The isotope ${}_{5}^{12}B$ having a mass $12.014 \text{ u}$ undergoes $\beta$-decay to ${}_{6}^{12}C$. ${}_{6}^{12}C$ has an excited state of the nucleus $({}_{6}^{12}C^*)$ at $4.041 \text{ MeV}$ above its ground state. If ${}_{5}^{12}B$ decays to ${}_{6}^{12}C^*$, the maximum kinetic energy of the $\beta$-particle in units of $\text{MeV}$ is ($1 \text{ u} = 931.5 \text{ MeV}/c^2$, where $c$ is the speed of light in vacuum).