COMEDK · Physics · 29. Nuclear Physics
In a deuterium-tritium fusion reaction given by \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_2^4 \mathrm{He}+{ }_0^1 n\) the mass defect is given as 0.0188 u . The total energy released in the fusion reaction is:
- A 2 MeV
- B 46 MeV
- C 188 MeV
- D 17.5 MeV
Answer & Solution
Correct Answer
(D) 17.5 MeV
Step-by-step Solution
Detailed explanation
The energy released in a nuclear reaction is calculated using the mass defect \(\Delta m\) and the conversion factor \(1 \text{ u} = 931.5 \text{ MeV}/c^2\). Given the mass defect \(\Delta m = 0.0188 \text{ u}\). The energy released \(Q\) is given by…
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