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Fusion bombs use neutrons from their fission trigger to create tritium fuel in the reaction $n + {}^{6}\textrm{Li} \to {}^{3}\textrm{H} + {}^{4}\textrm{He}$. What is the energy released by this reaction in MeV?
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$4.781 \textrm{ MeV}$
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OpenStax College Physics for AP® Courses Solution, Chapter 32, Problem 53 (Problems & Exercises) (1:18)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We are going to calculate the energy released when neutrons from a fission trigger hit lithium that's in the fusion part of the fusion bomb then that will create tritium and this will be what fuses in the fusion part of the reaction. So to calculate the energy released, we'll find the total mass of the reactants; there's neutron plus the mass of lithium-6 and then we'll take away from that the total mass of tritium and helium-4. And so we look up each of these masses in our appendix A and we have equal numbers of electrons surrounding the lithium and around the tritium and helium so these atomic masses even though they include electron masses, we don't need to worry about it because we have as many added to here as we are taking away amongst these two. So when we take the difference in the total number of atomic mass units of the reactants and the products and then multiply that by 931.5 megaelectron volts per c squared for every atomic mass unit times c squared we get an energy released of 4.781 megaelectron volts.