The nuclear fusion reaction is given as _1H^2 | vedclass

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(D) The given reaction is $_1^2H + _1^2H 	o _2^3He + n + 3.2 \, MeV$.The number of deuterium atoms in $2 \, kg$ $(2000 \, g)$ is calculated using the

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$10^{33} \, eV$

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(D) The given reaction is $_1^2H + _1^2H 	o _2^3He + n + 3.2 \, MeV$.The number of deuterium atoms in $2 \, kg$ $(2000 \, g)$ is calculated using the Avogadro number $(N_A \approx 6 	imes 10^{23} \, mol^{-1})$ and the molar mass of deuterium $(M_W = 2 \, g/mol)$:Number of atoms $= \frac{2000 \, g}{2 \, g/mol} 	imes 6 	imes 10^{23} \, mol^{-1} = 6 	imes 10^{26}$ atoms.Since $2$ deuterium atoms release $3.2 \, MeV$ of energy,the energy released per deuterium atom is $\frac{3.2}{2} = 1.6 \, MeV$.Total energy released for $6 	imes 10^{26}$ atoms $= 1.6 \, MeV 	imes 6 	imes 10^{26} = 9.6 	imes 10^{26} \, MeV$.Converting $MeV$ to $eV$ $(1 \, MeV = 10^6 \, eV)$:Total energy $\approx 10 	imes 10^{26} 	imes 10^6 \, eV = 10^{33} \, eV$.

The nuclear fusion reaction is given as $_1H^2 + _1H^2 \to _2He^3 + n + 3.2 \, MeV$. How much energy is released when $2 \, kg$ of deuterium undergoes fusion?

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