Question

Name the following nuclear reaction:
(a) ${}_{92}^{235}U+{}_0^{1}n\to {}_{38}^{90}Sr+{}_{54}^{143}Xe+3_0^{1}n+Y$
(b) ${}_1^{3}H+{}_1^{2}H\to \underset{2}{4}He+{}_0^{1}n+Y$

Answer 1

(a) ${}_{92}^{235}U+{}_0^{1}n\to {}_{38}^{90}Sr+{}_{54}^{143}Xe+3_0^{1}n+Y$

In the given nuclear reaction, the slow neutron strikes the parent heavy nucleus $\left({}_{92}^{235}U\right)$ and,splits into,two nearly equal fragments ${}_{38}^{90}\text{Sr}$ and ${}_{54}^{143}Xe$ with a release of 3 neutrons and a tremendous amount of energy. So this is a nuclear fission reaction.

(b) ${}_1^{3}H+{}_1^{2}H\to {}_2^{4}He+{}_0^{1}n+Y$

In the given nuclear reaction, the two light nuclei, i.e. tritium ${(}_1^{3}H)$ and deuterium ${(}_1^{2}H)$ combine to form a heavy helium nucleus ${(}_2^{4}He)$ with a release of 1 neutron and a tremendous amount of energy. So this is a nuclear fusion reaction.