ExamBro
ExamBro
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 \({ }_2 \mathrm{He}^4=7.0 \mathrm{MeV}\))

  1. A 8.1 MeV
  2. B 5.9 MeV
  3. C 23.6 MeV
  4. D 26.8 MeV
Verified Solution

Answer & Solution

Correct Answer

(C) 23.6 MeV

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & E_B=\mathrm{BE}_{\text {reactant }}-\mathrm{BE}_{\text {product }} \\ & =1.1 \times 2+1.1 \times 2-7 \times 4=-23.6 \mathrm{MeV} \\ & =\mathrm{Q}=23.6 \mathrm{MeV}\end{aligned}\)
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app