Verify that the total number of nucleons, total charge, and electron family number are conserved for each of the fusion reactions in the proton-proton cycle in {}^1 H + {}^1 H \rightarrow {}^2 H + e^+ + \nu_e {}^1 H + {}^2 H \rightarrow {}^3 He + \gamma {}^3 He + {}^3 He \rightarrow {}^4 He + {}^1 H + {}^1 H

Name: OpenStax College Physics Solution, Chapter 32, Problem 26 (Problems and Exercises)
Uploaded: 2021-05-21T16:30:23-07:00
Duration: 191 s
Description: Verify that the total number of nucleons, total charge, and electron family number are conserved for each of the fusion reactions in the proton-proton cycle in {}^1 H + {}^1 H \rightarrow {}^2 H + e^+ + \nu_e {}^1 H + {}^2 H \rightarrow {}^3 He + \gamma {}^3 He + {}^3 He \rightarrow {}^4 He + {}^1 H + {}^1 H

OpenStax College Physics for AP® Courses, Chapter 32, Problem 26 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. We are going to check to make sure that the total number of nucleons, the total charge and the total electron family number are conserved in each step of this proton-proton fusion cycle. So here's the first step where we have a proton plus another proton combining to form a deuterium nuclide plus a positron plus an electron-neutrino. So this nuclide here has one nucleon— just one proton— plus another single proton and that turns into two protons plus zero nucleons in this positron and zero nucleons in this electron-neutrino so we have two on the left and two on the right so that checks out. For charge, we have a positive one here and a positive one for this proton for a total of two and that turns into a positive one in this deuterium nuclide plus a positive one in this positron and zero charge for the electron-neutrino so it's two on both sides so that checks out. And for the electron family number neither of these protons have an electron family number nor does the deuterium nuclide. The positron has an electron family number of negative 1 and the electron-neutrino has an electron family number of positive 1 so that also makes zero in total on the right-hand side so there we go! The next step is that a proton combines with a deuterium nuclide to form helium-3 plus a gamma ray. So there's one nucleon in the proton, there's two nucleons in the deuterium nuclide because there's one proton plus one neutron so that's a total of three nucleons on the left and on the right hand side, we have 3 nucleons in this helium nuclide— two protons plus one neutron— and zero nucleons in the gamma particle. So that is 3 on both sides—that checks out— and we have positive 1, positive 1 in each of the protons on the left and then we have positive 2 among the two protons in the helium nuclide so that's 2 in total on both sides for the charge and then the electron family number is 0 everywhere here each of these particles has 0 for electron family number so it's zero on the left and on the right. And the last step we have helium-3 plus a helium-3 nuclide combining to form a helium-4 nuclide plus 2 protons. So there are 3 nucleons in this helium-3 nuclide plus 3 nucleons in this one as well for a total of 6 on the left and then these 4 nucleons here— 2 protons plus 2 neutrons— plus 1 nucleon here in this proton plus 1 more here for a total of 6 on the left and 6 on the right. There is a charge of 2 for the two protons in this helium nuclide plus 2 here and then 2 in the helium nuclide on the right plus 1 in each of these protons for a total of 4 on the right and 4 on the left and none of these particles in this step have an electron family number so it's 0 on both sides.