Question

The energy of the reaction ${Li}^7+p⟶2{He}^4$ is (the binding energy per nucleon in ${Li}^7$ and ${He}^4$ nuclei are 5.60 and 7.06 MeV respectively.)

• A. 17.3 MeV`
• B. 1.73 MeV
• C. 1.46 MeV
• D. Depends on binding energy of proton

Answer 1

The correct option is D Depends on binding energy of proton

The energy released will depend on the energy equivalent of proton, taking its mass to be $1amu$ or

corresponding energy as $931.5Mev$

we get energy released as $Q=E_{reactant}-E_{product}=7\times 5.6+931.5-2\times 4\times 7.06=914.22Mev$