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The binding energy per nucleon of ^7_3Li and ^4_2He nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction ^7_3Li + ^1_1H → ^4_2He + Q, the value of energy Q released is:

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

Correct answer: 17.3 MeV

Solution

The energy released (Q) is calculated using the difference in total binding energy of the products and reactants. For ^7_3Li, the total binding energy is 7 × 5.60 = 39.2 MeV. For ^4_2He, it is 4 × 7.06 = 28.24 MeV. The total binding energy of the products is 28.24 MeV (from ^4_2He). The difference, Q = 28.24 - 39.2 + 7.06 = 17.3 MeV.

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