Radioactivity — Question 11

(a) A nuclear fusion reaction is given below:

$\begin{matrix} _{1}^{2}\text{H} & + & _{1}^{2}\text{H} & \longrightarrow & _{2}^{3}\text{He} & + & _{0}^{1}\text{n} & + & 3.3\text{ MeV} \\ \footnotesize{\text{(deuterium)}} & & \footnotesize{\text{(deuterium)}} & & \footnotesize{\text{(helium isotope)}} & & \footnotesize{\text{(neutron)}} \end{matrix}$

$\begin{matrix} _{2}^{3}\text{He} & + & _{1}^{2}\text{H} & \longrightarrow & _{2}^{4}\text{He} & + & _{1}^{1}\text{H} & + & 18.3\text{ MeV} \\ \footnotesize{\text{(helium isotope)}} & & \footnotesize{\text{(deuterium)}} & & \footnotesize{\text{(helium)}} & & \footnotesize{\text{(proton)}} \end{matrix}$

In this reaction, three deuterium nuclei fuse to form a helium nucleus.

(b) Approximately, 21.6 MeV energy is released in the reaction.

(c) When two deuterium nuclei fuse, 3.3 MeV energy is released and the nucleus of helium isotope ($_{2}^{3}\text{He}$ ) is formed. This helium again gets fused with one deuterium nucleus to form a helium nucleus ($_{2}^{4}\text{He}$) and 18.3 MeV is released.