Final answer:
The energy released by the nuclear reaction A + B → C + D with the given masses is 19551 MeV, calculated using the mass-energy equivalence where a mass defect of 21 amu is converted to energy by multiplying with 931 MeV/amu.
Step-by-step explanation:
To calculate the energy released by the nuclear reaction A + B → C + D, where the nuclear masses of A, B, C, D are 1, 63, 3, 40 amu respectively, we need to use the relationship that 1 amu equals 931 MeV.
The mass defect (Δm) is the difference between the total mass of the reactants and the total mass of the products, which in this case is (1 + 63) - (3 + 40) = 21 amu.
The energy released (E) is then given by E = Δm × 931 MeV/amu.
Therefore, the energy released by this reaction is 21 amu × 931 MeV/amu = 19551 MeV.