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Question

Calculate the energy that can be obtained from 1 kg of water through the fusion reaction
2H + 2H → 3H + p.
Assume that 1.5 × 10−2% of natural water is heavy water D2O (by number of molecules) and all the deuterium is used for fusion.

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Solution

Given:
18 g of water contains 6.023 × 1023 molecules.

∴ 1000 g of water =6.023×1023×100018=3.346×1025 molecules
% of deuterium = 3.346×1025 × 0.015100 = 0.05019 × 1023
Energy of deuterium = 30.4486×1025
=2×mH2-mH3-mpc2=2×2.014102 u-3.016049 u-1.007276 uc2=0.004879×931 MeV=4.542349 MeV=7.262 ×10-13 J
Total energy = 0.05019 × 1023 × 7.262 × 10-13 J
= 3644 MJ

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