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Verify that the fusion of 1.0 $\mathrm{kg}$ of deuterium by the reaction $$

^{2} \mathrm{H}+^{2} \mathrm{H} \rightarrow^{3} \mathrm{Hc}+\mathrm{n} \quad(Q=+3.27 \mathrm{MeV})$$

$2.5 \times 10^{4} \mathrm{y}$

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Cornell University

University of Michigan - Ann Arbor

University of Washington

McMaster University

All right, So, uh, in this problem, we have the reaction. Ah, the DEA Terry, um, becomes hit him three and plus when you troll. So, uh, the cue for this reaction is ah, three 0.21. Maggie. So, um, if we have ah, one kilograms off the deal could Ah, deuterium. So we want to know. Ah, the energy that we released. So, uh, if we have one kilograms off the deuterium, then the total number of particle will be 1000 grams over Ah, two times. Sorry. Just the two. Just two grams per mohs. And if we want to know how many? Ah, him three will be created then when we need to develop another two. So this better is to 50 most. And we can transfer this into ah, the number of particles that it would be. Ah. So just use 2 50 miles. Sorry. Just used to 50 miles. Times 6102 times 10 to the 23rd. Ah, promo. So the battery's ah, let me see 1.5 times 10 to the 26. So basically we have one kilogram self that do you carry him? We will have finally ate 1.5 times. 10 to 26 him three will be created. Um, and we also know the cue for each reaction. So the total energy will be you equal to Q times. So this supposed is equal to end. Cute. I'm saying so. The total energy is, uh, embassy seven point I times 10 to the the teens, Jos. Okay. And a weekend? Because supposed we have Ah, we have a bump. We have Ah! Ah. We have bump was power 100 warts, so we can see the total time for lighting will be. You overwhelm 100 warts. So it equal 7.9 times 10 to 11 seconds. Oh, if you're transferring two years, this is 2.5 times. 10 to the fourth years, okay?