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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Physics
Chapter
Nuclear Physics

JEE Main 2025 · 2 April · Evening Shift · Physics · Nuclear Physics

Energy released when two deuterons ( _1 H^2) fuse to form a helium...

Energy released when two deuterons (1H2)\left({ }_1 \mathrm{H}^2\right) fuse to form a helium nucleus (2He4)\left({ }_2 \mathrm{He}^4\right) is :
(Given : Binding energy per nucleon of 1H2=1.1MeV{ }_1 \mathrm{H}^2=1.1 \mathrm{MeV} and binding energy per nucleon of 2He4=7.0MeV{ }_2 \mathrm{He}^4=7.0 \mathrm{MeV})

Options

  1. A8.1 MeV
  2. B5.9 MeV
  3. C23.6 MeV
  4. D26.8 MeV

Correct answer

C. 23.6 MeV

Step-by-step solution

EB=BEreactant −BEproduct =1.1×2+1.1×2−7×4=−23.6MeV=Q=23.6MeV\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}
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