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.

Answer 1

Given:

18 g of water contains 6.023 $\times$ 10²³ molecules.

∴ 1000 g of water =$\frac{6.023\times {10}^{23}\times 1000}{18}=3.346\times {10}^{25}$ molecules
% of deuterium = $3.346\times {10}^{25}$ $\times$ $\frac{0.015}{100}$ = 0.05019 $\times$ 10²³
Energy of deuterium = $30.4486\times {10}^{25}$
$$\begin{gathered} =\left[2\times m\left({}^{2}H\right)-m\left({}^{3}H\right)-m_p\right]c^2 \\ =\left(2\times 2.014102\mathrm{u}-3.016049\mathrm{u}-1.007276u\right)c^2 \\ =0.004879\times 931\mathrm{MeV} \\ =4.542349\mathrm{MeV} \\ =7.262\times {10}^{-13}\mathrm{J} \end{gathered}$$
Total energy = 0.05019 $\times$ 10²³ $\times$ 7.262 $\times$ 10^$-$13 J
= 3644 MJ