Question

The binding energy per nucleon of a Deuteron $\left({}_1^{2}H\right)$ and helium nucleus $\left({}_2^{4}He\right)$ is $1.1MeV$ and $7MeV$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is :

• A. $13.9MeV$
• B. $26.9MeV$
• C. $23.6MeV$
• D. $19.2MeV$

Answer 1

Answer: (C)

$23.6MeV$

$\begin{array}{c}\text{* Given-}\ast \text{Two deutron reacts to form a Helium nucleus.}\\ \text{* BE per nucleon of Deutron = 1.1MeV and Helium = 7Mev}\end{array}$

${}_1^{2}H{+}_1^{2}H\to {}_2^{4}He$

$\begin{array}{c}\text{BE = Binding energy per nucleons}\text{X number of Nucleons}\\ BE=BE\text{product}-B{E}_{\text{Reactant}}\\ BE=[7\times 4]-[1.1\times 2\times 2]=28meV-4.4MeV\\ BE=23.6MeV\end{array}$