Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 32 - Elementary Particles - Search and Learn - Page 946: 3

Answer

\begin{array}{|c|c|c|c|c|c|} \hline \rm{Decay}& \rm Spin & {\rm Mass/Energy(MeV/ }c^2)&{\rm Charge} (e)&\rm Baryon \;number&\rm Lepton \;number\\ \hline \rm t\rightarrow W^++b& \frac{1}{2}=1- \frac{1}{2}& 137500\gt\overbrace{80385+4180}^{84565}& \frac{2}{3}=1-\frac{1}{3}&\frac{1}{3}=0+\frac{1}{3}&0=0+0\\ \hline \rm W^+\rightarrow u+\overline{d}&1= \frac{1}{2}+ \frac{1}{2}& 80385\gt\overbrace{2.3+4.8}^{7.1}& 1=\frac{2}{3}+\frac{1}{3}&0=\frac{1}{3}-\frac{1}{3}&0=0+0\\ \hline \rm \overline{t}\rightarrow W^-+\overline{b}& \frac{1}{2}=1- \frac{1}{2}& 137500\gt\overbrace{80385+4180}^{84565}& -\frac{2}{3}=-1+\frac{1}{3}&-\frac{1}{3}=0-\frac{1}{3}&0=0+0\\ \hline \rm W^-\rightarrow \mu^-+\overline{\nu_\mu}&1= \frac{1}{2}+ \frac{1}{2}& 80385\gt\overbrace{105.7+0.00014}^{105.7}& -1=-1+0 &0=0+0 &0=1-1\\ \hline \end{array}

Work Step by Step

In the table below, if the sum of spin is equal, so the spin is conserved. And the same for charge, Beryon number, and Lepton number. But for Mass-energy, the mass of the daughters must be less than the mass of the initial particle, as you see in the table below. \begin{array}{|c|c|c|c|c|c|} \hline \rm{Decay}& \rm Spin & {\rm Mass/Energy(MeV/ }c^2)&{\rm Charge} (e)&\rm Baryon \;number&\rm Lepton \;number\\ \hline \rm t\rightarrow W^++b& \frac{1}{2}=1- \frac{1}{2}& 137500\gt\overbrace{80385+4180}^{84565}& \frac{2}{3}=1-\frac{1}{3}&\frac{1}{3}=0+\frac{1}{3}&0=0+0\\ \hline \rm W^+\rightarrow u+\overline{d}&1= \frac{1}{2}+ \frac{1}{2}& 80385\gt\overbrace{2.3+4.8}^{7.1}& 1=\frac{2}{3}+\frac{1}{3}&0=\frac{1}{3}-\frac{1}{3}&0=0+0\\ \hline \rm \overline{t}\rightarrow W^-+\overline{b}& \frac{1}{2}=1- \frac{1}{2}& 137500\gt\overbrace{80385+4180}^{84565}& -\frac{2}{3}=-1+\frac{1}{3}&-\frac{1}{3}=0-\frac{1}{3}&0=0+0\\ \hline \rm W^-\rightarrow \mu^-+\overline{\nu_\mu}&1= \frac{1}{2}+ \frac{1}{2}& 80385\gt\overbrace{105.7+0.00014}^{105.7}& -1=-1+0 &0=0+0 &0=1-1\\ \hline \end{array} It is obvious now that all decays of the given figure are conserved in spin, mass-energy, charge, Baryon number, and Lepton number.
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