When a slow neutron goes sufficiently close to a ${U^{235}}$ nucleus,then the process that takes place is

Fast neutrons can easily be slowed down by

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).

If there is a mass defect of $0.1\%$ in a nuclear fission process,how much energy will be released in the fission of $1\, kg$ of mass?

In a nuclear reactor,heavy nuclei are not used as moderators because:

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.