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The binding energy per nucleon for the nuclei ⁷₃Li and ⁴₂He is 5.60 MeV and 7.06 MeV, respectively. For the nuclear reaction p + ⁷₃Li arrow 2 ⁴₂He, the proton must have an energy of

1. 28.24 MeV
2. 17.28 MeV
3. 1.46 MeV
4. 39.2 MeV

Correct answer: 17.28 MeV

Solution

Total binding energy of products = 2 x (4 x 7.06) = 56.48 MeV; reactant = 7 x 5.60 = 39.2 MeV (proton has no binding energy). Energy required/released = 56.48 - 39.2 = 17.28 MeV.

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