Calculate the energy released in the following fusion reaction where reactants are ⁶Li and a neutron;products are ⁴He and ³H.(atomic masses: ⁶Li,6.015 12;neutron,1.008 67;⁴He,4.002 60;³H,3.016 031;also 1 u = 931.5 MeV/c²) A)2.95 MeV B)4.81 MeV C)8.63 MeV D)17.2 MeV

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Quizplus · Accepted Answer

The energy released in a nuclear reaction can be calculated using the mass defect and converting it into energy using Einstein's equation $E = \Delta m c^2$, where $\Delta m$ is the mass defect and $c$ is the speed of light. However, in nuclear physics problems, it's common to use atomic mass units (u) and directly convert mass defect into energy using the conversion factor $1 u = 931.5 \text{ MeV/c}^2$.Given atomic masses:- $^7Li = 6.015 12 \, u$- $n = 1.008 67 \, u$- $^4He = 4.002 60 \, u$- $^3H = 3.016 031 \, u$The total mass of reactants is $6.015 12 \, u + 1.008 67 \, u = 7.023 79 \, u$.The total mass of products is $4.002 60 \, u + 3.016 031 \, u = 7.018 631 \, u$.The mass defect ($\Delta m$) is the difference between the mass of the reactants and the mass of the products: $7.023 79 \, u - 7.018 631 \, u = 0.005 159 \, u$.Converting this mass defect into energy: $0.005 159 \, u \times 931.5 \, \text{MeV/c}^2 = 4.81 \, \text{MeV}$.Therefore, the energy released in this fusion reaction is 4.81 MeV.

Calculate the Energy Released in the Following Fusion Reaction Where