Problem: The positions of 2 4 He , 1 2 D and 3 7 L i are shown on the binding-energy curve (as in the figure). The energy released in the fusion reaction 1 2 D + 3 7 L i → 2 4 He + 2 4 H

Question

Problem: The positions of 2 4 ​ He , 1 2 ​ D and 3 7 ​ L i are shown on the binding-energy curve (as in the figure). The energy released in the fusion reaction 1 2 ​ D + 3 7 ​ L i → 2 4 ​ He + 2 4 ​ He + 0 1 ​ n will be closest to:

TestHub · Accepted Answer

The energy released in a nuclear reaction can be determined from the difference in binding energies of the reactants and products. The binding energy per nucleon can be read off the graph.

$_{1}^{2} D$ : 1 MeV/nucleon

$_{3}^{7} L i$ : Approximately 5.5 MeV/nucleon

$_{2}^{4} He$ : 7 MeV/nucleon

Calculate the total binding energy for each nucleus:

$_{1}^{2} D$ : 2 nucleons × 1 MeV/nucleon = 2 MeV

$_{3}^{7} L i$ : 7 nucleons × 5.5 MeV/nucleon = 38.5 MeV

$_{2}^{4} He$ : 4 nucleons × 7 MeV/nucleon = 28 MeV

Total Binding Energy of Reactants:

$2 + 38.5 = 40.5, M e V$

Total Binding Energy of Products:

$28 + 28 + 0 = 56 M e V$

$Q = 56 - 40.5 = 15.5 M e V$

The energy released in the fusion reaction is approximately 15.5 MeV, which is closest to 16 MeV.

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