Question 187 of 239

Energy released when two deuterons $\left({ }_1 \mathrm{H}^2\right)$ fuse to form a helium nucleus $\left({ }_2 \mathrm{He}^4\right)$ is :
(Given : Binding energy per nucleon of ${ }_1 \mathrm{H}^2=1.1 \mathrm{MeV}$ and binding energy per nucleon of ${ }_2 \mathrm{He}^4=7.0 \mathrm{MeV}$)

A8.1 MeVB5.9 MeVC23.6 MeVD26.8 MeV

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Answer: (C) 23.6 MeV

$\begin{aligned} & E_B=\mathrm{BE}_{\text {reactant }}-\mathrm{BE}_{\text {product }} \\ & =1.1 \times 2+1.1 \times 2-7 \times 4=-23.6 \mathrm{MeV} \\ & =\mathrm{Q}=23.6 \mathrm{MeV}\end{aligned}$