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# $\bullet$ In a hypothetical nuclear-fusion reactor, two deuteriumnuclei combine or "fuse" to form one helium nucleus. Themass of a deuterium nucleus, expressed in atomic mass units(u), is 2.0136 u; that of a helium nucleus is 4.0015 u.$\left(1 \mathrm{u}=1.661 \times 10^{-27} \mathrm{kg} .\right)$ (a) How much energy is releasedwhen 1.0 $\mathrm{kg}$ of deuterium undergoes fusion? (b) The annualconsumption of electrical energy in the United States is on theorder of $1.0 \times 10^{19} \mathrm{J} .$ How much deuterium must react to pro-duce this much energy?

## 17421 $\mathrm{kg}$

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all right and part of this problem. We want to know how much energy Khun b released from converting a kilogram of deuterium into helium. So the first thing we need to do is we need to figure out how many dude Iran's there are in one kilogram and were given that one. Dude, Iran has a ah atomic mass of 2.136 atomic mass units. And the conversion for this two kilograms is 1.661 times 10 to the negative 27. And this comes out to 2.99 um, times 10 to the 26. Then if we if this gets converted to helium, we end up with half the number of helium atoms because it takes to do drones to form one helium. So this is the number of peeling nuclear weekend, and the mass associated without many nuclei is gonna be this number times the way of a single one. And we were given that their masses for point oo 15 atomic last units and again, we want to convert this two kilograms. Oh, no. This comes out to 0.993 65 kilograms. So the amount that we lost in the reaction is one minus that, and we multiply this by C squared to get 5.7 times tended 14 Jules and in part B. We want to figure out, um, how many kilograms it would take. Two power. The U. S. So that's tender. The 19 Jules divided by a 5.7 times 10 to the 14 Jules per kilogram. And this equals 1.8 times 10 to the four kilograms, which is not very much at all.

University of Hawaii at Manoa

Gravitation

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