- $\textrm{U} \to \textrm{Ra} + \textrm{He}$
- $\textrm{Ra} \to \textrm{Pb} + \textrm{C}$
- $\textrm{C} \to \textrm{N} + e^- + \bar{\nu_e}$
- $\textrm{Mg} \to \textrm{Na} + e^+ + \nu_e$

- No, because of conservation of charge violation.
- Yes
- Yes
- Yes

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View sample solutionThis is College Physics Answers with Shaun Dychko. Let's determine whether each of these following reactions are possible based on violation of conservation laws or not. So in option (a), we have hypothetically uranium decaying into radium plus helium. And so the charge on uranium is 92 protons and radium has a charge of 88 and helium has 2 protons and 88 plus 2 is 90 and we started with 92 so this would suggest some charge is lost so that's a violation of the conservation of charge; this decay is not possible. Part (b) we have radium decaying into lead plus carbon so let's test whether the charge is conserved. So we have 88 protons initially; lead has 82 and carbon has 6 so 82 and 6 is also 88 and so yes, according to conservation of charge, this is okay. And we can look at the number of nucleons and if you look at appendix A, radium has 226 nucleons and there is an isotope of lead and an isotope of carbon that when added together also makes 226 and so we can say, yes this is possible and it would end up making the rare isotope of carbon-14. Part (c) we have carbon turning into nitrogen plus a <i>β</i>-particle which is an electron and an electron anti-neutrino. So looking at charge, we have 6 protons in the carbon nucleus and then in nitrogen, we have 7 protons so that's an additional positive charge but it's compensated for by a negative electron emitted. So basically a neutron in the carbon nucleus has turned into a proton and an electron and charge is conserved because 7 plus negative 1 is still 6 the anti-neutrino having no charge. The number of nucleons, well, if this is carbon-14 which is the isotope of carbon that decays so is 14 nucleons; nitrogen also has 14 nucleons and these are not nucleons, either of those so they are 0 and 0 so we have 14 equals 14 so that's good and then we can test the electron family number. So this is a nucleus and so it has an electron family number of 0 and same with nitrogen and then the <i>β</i>-particle has an electron family number of negative 1 whereas the anti-neutrino has a electron family number of positive 1 and so this works out to zero. OK. So yes this is possible because it has violated no conservation laws. And then part (d), magnesium turning into well I should return up here and correct this little thing. The regular matter has an electron family number of positive and the anti-matter has an electron family number that is negative. OK. Either way, the sum is still zero. Okay, down here magnesium to sodium. Charge we have 12 protons in magnesium and 11 in sodium but we have an additional 1 in this positron and this makes a total of 12 so that's good. Number of nucleons is 24 in the magnesium and 24 in the sodium so that checks out. And then the electron family number, we have negative 1 for this electron anti-particle also called a positron, and we have an electron family number of positive 1 for this electron anti-neutrino giving a total of 0 on the right-hand side matching 0 on the left and so yes, this is possible.