In a deuterium-tritium fusion reaction given by ₁^2 H + ₁^3 H ₂^4 He + ₀^1 n the mass defect is given as 0.0188 u . The total energy released in the fusion reaction is:

Step-by-step solution

The energy released in a nuclear reaction is calculated using the mass defect m and the conversion factor 1 u = 931.5 MeV /c^2 . Given the mass defect m = 0.0188 u . The energy released Q is given by Q = m 931.5 MeV / u . Q = 0.0188 931.5 MeV . Q 17.5122 MeV . Rounding to the nearest provided option, the energy released is 17.5 MeV . Answer: 17.5 MeV