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If the binding energy per nucleon in ³Li⁷ and ²He⁴ nuclei are respectively 5.60 MeV and 7.06 MeV, then the energy of proton in the reaction ³Li⁷ + p → ²He⁴ is:

1. 19.6 MeV
2. 2.4 MeV
3. 8.4 MeV
4. 17.3 MeV

Correct answer: 17.3 MeV

Solution

The total binding energy of ³Li⁷ is 7 × 5.60 = 39.2 MeV, and for ²He⁴, it is 4 × 7.06 = 28.24 MeV. The energy released in the reaction is the difference in binding energy: 39.2 - 28.24 = 10.96 MeV. Adding the proton's binding energy (7.06 MeV), the total energy is approximately 17.3 MeV.

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