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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.
(Use Mass of proton mp = 1.007276 u, Mass of `""_1^1"H"` atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)
Solution
18 g of water contains `6.023 xx 10^23` molecules.
`∴ "1000 g of water" = (6.023 xx 10^23 xx 1000)/18 = 3.346 xx 10^25` molecules
`% "of deuterium" = 3.346 xx 10^25 xx 0.015/100 = 0.05019 xx 10^23`
Energy of deuterium = `30.4486 xx 10^25`
`= [2 xx m(""^2H) - m(""^3H) - m_p]c^2`
`= (2 xx 2.014102 "u" - 3.016049 "u" - 1.007276 "u")c^2`
`= 0.004879 xx 931 "MeV"`
`= 4.542349" MeV"`
`= 7.262 xx 10^-13 "J"`
Total energy = `0.05019 xx 10^23 xx 7.262 xx 10^-13 "J"`
= 3644 MJ
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