Energy released when two deuterons $\left({ }_1 \mathrm{H}^2\right)$ fuse to form a helium nucleus $\left({…

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 ${ }_2 \mathrm{He}^4=7.0 \mathrm{MeV}$)
  1. 8.1 MeV
  2. 5.9 MeV
  3. 23.6 MeV
  4. 26.8 MeV

Solution


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

Asked in: JEE Main 2025 (02 Apr Shift 2)

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