Question

The binding energy of deuteron is 2.2 MeV and that of ${}_2^{4}He$ is 28 MeV. If two deutron are fused to form one ${}_2^{4}He$ then the energy released is.

• A. $30.2MeV$
• B. $25.8MeV$
• C. $23.6MeV$
• D. $19.2MeV$

Answer 1

The correct option is C $23.6MeV$

$\begin{array}{c}\text{Binding energy: Energy released due to formation}\\ \text{of atom by constituent particle or energy}\\ \text{nedded to break atom in constituent}\\ \text{particle.}\end{array}$

$\begin{array}{cc}BE\text{of}\mathrm{deutron}\left({}_1^{2}H\right)& =2.2MeV\\ BE\text{of}{}_2^{4}He=28Mev& \end{array}$

$\begin{array}{c}when\\ {}_1^{2}H{+}_1^{2}H{\to }_2^{4}H\end{array}$

$\begin{array}{c}\text{Energy released}=\text{Energy of product - Reactemt}\\ \begin{array}{cc}& =28-2\times 2.2\\ & =28-4.4\\ & =23.6MeV\end{array}\end{array}$