Energy released in the fusion of 1, kg of deu | vedclass

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(A) The fusion reaction of deuterium is given by: $_1H^2 + _1H^2 	o _2He^3 + _0n^1 + 3.27\,MeV$.In $1\, kg$ of deuterium $(^2H)$,the number of atoms

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$8 	imes 10^{13}\,J$

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(A) The fusion reaction of deuterium is given by: $_1H^2 + _1H^2 	o _2He^3 + _0n^1 + 3.27\,MeV$.In $1\, kg$ of deuterium $(^2H)$,the number of atoms is $N = \frac{1000}{2} 	imes 6.022 	imes 10^{23} = 3.011 	imes 10^{26}$ atoms.Since each fusion reaction involves two deuterium nuclei,the number of fusion events is $N_{fusion} = \frac{N}{2} = 1.5055 	imes 10^{26}$.The energy released per reaction is $3.27\,MeV = 3.27 	imes 1.602 	imes 10^{-13}\,J \approx 5.238 	imes 10^{-13}\,J$.Total energy $E = N_{fusion} 	imes 5.238 	imes 10^{-13}\,J = 1.5055 	imes 10^{26} 	imes 5.238 	imes 10^{-13} \approx 7.88 	imes 10^{13}\,J$.Rounding this gives approximately $8 	imes 10^{13}\,J$.

Energy released in the fusion of $1\, kg$ of deuterium nuclei is:

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