Question
(a) Show that the conjectured decay of the proton, p→π0+e+p \rightarrow \pi^0 + e^+, violates conservation of baryon number and conservation of lepton number. (b) What is the analogous decay process for the antiproton?
Question by OpenStax is licensed under CC BY 4.0
Final Answer
  1. Please see the solution video.
  2. p‾→π0+e−\overline{p} \rightarrow \pi^0 + e^-

Solution video

OpenStax College Physics, Chapter 33, Problem 34 (Problems & Exercises)

OpenStax College Physics, Chapter 33, Problem 34 (PE) video thumbnail

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Video Transcript
This is College Physics Answers with Shaun Dychko. A proton decaying into a neutral pion and a positron violates the conservation of baryon number. When we consult table [33.2], we see the proton has a baryon number of 1 and the neutral pion has a baryon number of 0 and the electron or positron has a baryon number of 0 as well so it's 1 on the left and total of 0 on the right and so baryon number is not conserved. The electron family number is also not conserved because the proton has a 0 for the electron family number whereas the electron or positron has negative 1 and the neutral pion has 0 and this should be negative 1 there so electron family number is not conserved. This is the lepton number of concern here; there's also the tau and muon lepton family numbers but they don't matter. Okay! Part (b) asks us to write the analogous decay for the anti-proton and so we take the anti-particle for each of these and the neutral pion is strange because its anti-particle is itself and so I don't need to write a bar on top of the neutral pion and the anti-particle to the positron is the electron.