Question
The natural abundance of deuterium in water is $0.0156 \%$ (i.e., $0.0156 \%$ of the hydrogen nuclei in water are ${ }^{2} \mathrm{H}$ ). If the fusion reaction $\left({ }^{2} \mathrm{H}+{ }^{2} \mathrm{H}\right)$ yields $3.65 \mathrm{MeV}$ of energy on average, how much energy could you get from $1.00 \mathrm{L}$ of water? (There are two reactions with approximately equal probabilities; one yields $4.03 \mathrm{MeV}$ and the other 3.27 MeV.) Assume that you are able to extract and fuse $87.0 \%$ of the deuterium in the water. Give your answer in kilowatt hours.
Step 1
Given that the natural abundance of deuterium in water is $0.0156 \%$, the amount of deuterium in 1 liter (or 1000 g) of water is: \[M = 0.0156 \% \times M_W = \frac{0.0156}{100} \times 1000 \, \text{g} = 0.156 \, \text{g}\] Show more…
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A water sample is found to have $0.016 \%$ deuterium content (that is, $0.016 \%$ of the hydrogen nuclei in the water are $^{2} \mathrm{H}$ ). If the fusion reaction $\left(^{2} \mathrm{H}+^{2} \mathrm{H}\right)$ yields 3.65 MeV of energy on average, how much energy could you get from $1.00 \mathrm{L}$ of the water? (There are two reactions with approximately equal probabilities; one yields $4.03 \mathrm{MeV}$ and the other $3.27 \mathrm{MeV} .$ ) Assume that you are able to extract and fuse $87.0 \%$ of the deuterium in the water. Give your answer in kilowatt hours.
Approximately 1 of every 3 300 water molecules contains one deuterium atom. (a) If all the deuterium nuclei in 1 $\mathrm{L}$ of water are fused in pairs according to the $\mathrm{D}-\mathrm{D}$ fusion reaction $^{2} \mathrm{H}+^{2} \mathrm{H} \rightarrow^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}$ , how much energy in joules is liberated? (b) What If? Burning gasoline produces approximately $3.40 \times 10^{7} \mathrm{J} / \mathrm{L}$ . State how the energy obtainable from the fusion of the deuterium in 1 $\mathrm{L}$ of water compares with the energy liberated from the burning of 1 $\mathrm{L}$ of gasoline.
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