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Verify by listing the number of nucleons, total charge, and electron family number before and after the cycle that these quantities are conserved in the overall proton-proton cycle in $2e^- + 4{}^{1}\textrm{H} \to {}^{4}\textrm{He} + 2\nu_e + 6\gamma$.
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OpenStax College Physics Solution, Chapter 32, Problem 29 (Problems & Exercises) (1:49)

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This is College Physics Answers with Shaun Dychko. The overall proton-proton fusion cycle involves four protons plus two electrons creating a single helium for two electron neutrino's and six gamma rays. We are going to verify that the number of nucleons, the electron family number and the charge are conserved in this reaction. So for the number of nucleons, we have zero when we are considering these two electrons and four of them considering these protons. And that's on the left side, we have a total of four then and on the right hand side, it's helium-4 and so it clearly has four nucleons and then there are no nucleons in the neutrino's or the gamma rays and so yes, this conservation law is valid and there's four on each side. Electron family number well there are two electrons each having an electron family number of one for a total of two and there are no electron family number attributed to protons and nor attributed to helium nuclei but there are two electron family numbers attributed to two electron neutrino's and zero for the gamma rays. And so yes we verified that electron family number is conserved; it is two on both sides. Now for charge, we have a negative 2 charge for these two electrons and a positive 4 charge for these four protons for a total of positive 2 net charge. And on the right hand side, 0 charge for the gamma rays, 0 charge for the neutrino's and 2 charge for the two protons in the helium nucleus. And so that net charge of 2 on the right side and the net charge of 2 on the left side means yes, the charge is conserved.