JEE Mains · Physics · STD 12 - 13. Nuclei
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 \({ }_2 \mathrm{He}^4=7.0 \mathrm{MeV}\))
- A 8.1 MeV
- B 5.9 MeV
- C 23.6 MeV
- D 26.8 MeV
Answer & Solution
Correct Answer
(C) 23.6 MeV
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & E_B=\mathrm{BE}_{\text {reactant }}-\mathrm{BE}_{\text {product }} \\ & =1.1 \times 2+1.1 \times 2-7 \times 4=-23.6 \mathrm{MeV} \\ & =\mathrm{Q}=23.6 \mathrm{MeV}\end{aligned}\)
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