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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 =$\frac{6.023×{10}^{23}×1000}{18}=3.346×{10}^{25}$ molecules % of deuterium = $3.346×{10}^{25}$ $×$ $\frac{0.015}{100}$ = 0.05019 $×$ 1023 Energy of deuterium = $30.4486×{10}^{25}$ $=\left[2×m\left({}^{2}H\right)-m\left({}^{3}H\right)-{m}_{p}\right]{c}^{2}\phantom{\rule{0ex}{0ex}}=\left(2×2.014102\mathrm{u}-3.016049\mathrm{u}-1.007276u\right){c}^{2}\phantom{\rule{0ex}{0ex}}=0.004879×931\mathrm{MeV}\phantom{\rule{0ex}{0ex}}=4.542349\mathrm{MeV}\phantom{\rule{0ex}{0ex}}=7.262×{10}^{-13}\mathrm{J}$ Total energy = 0.05019 $×$ 1023 $×$ 7.262 $×$ 10$-$13 J = 3644 MJ

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