COMEDK · Physics · 29. Nuclear Physics
If the binding energy per nucleon in \({ }_3 \mathrm{Li}^7\) and \({ }_2 \mathrm{He}^4\) nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of p in the reaction \(p+{ }_3 L i^7 \rightarrow 2{ }_2 H e^4\) is
- A 17.28 MeV
- B 28.28 MeV
- C 12.28 MeV
- D 13.28 MeV
Answer & Solution
Correct Answer
(A) 17.28 MeV
Step-by-step Solution
Detailed explanation
The given nuclear reaction is \(p + _{3}Li^{7} \rightarrow 2_{2}He^{4}\). The total binding energy of the reactants is: Binding energy of \(p\) = 0 MeV (since it is a single nucleon). Binding energy of \(_{3}Li^{7} = 7 \times 5.60 \text{ MeV} = 39.20 \text{ MeV}\). Total binding…
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