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\(^{238}_{92}A \rightarrow ^{234}_{90}B + ^{4}_{2}D + Q\) In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of \(^{238}_{92}A = 238.05079 \times 931.5\) MeV/c², mass of \(^{234}_{90}B = 234.04363 \times 931.5\) MeV/c², mass of \(^{4}_{2}D = 4.00260 \times 931.5\) MeV/c²]
- 3.82 MeV
- 5.9 MeV
- 2.12 MeV
- 4.25 MeV
Correct answer: 4.25 MeV
Solution
The correct option is 4.25 MeV because it represents the energy released during the reaction, calculated by finding the mass difference between the reactants and products and converting that mass defect into energy using Einstein's equation, E=mc².
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