At the Stanford Linear Accelerator, electrons and positrons collide together at very high energies to create other elementary particles. Suppose an electron and a positron, each with rest energies of \(0.511 \mathrm{MeV},\) collide to create a proton (rest energy \(938 \mathrm{MeV}\) ), an electrically neutral kaon \((498 \mathrm{MeV}),\) and a negatively charged sigma baryon \((1197 \mathrm{MeV}) .\) The reaction can be written as: $$\mathrm{e}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{p}^{+}+\mathrm{K}^{0}+\Sigma^{-}$$ (a) What is the minimum kinetic energy the electron and positron must have to make this reaction go? Assume they each have the same energy. (b) The sigma can decay in the reaction \(\Sigma^{-} \rightarrow \mathrm{n}+\pi^{-}\) with rest energies of \(940 \mathrm{MeV}\) (neutron) and \(140 \mathrm{MeV}\) (pion). What is the kinetic energy of each decay particle if the sigma decays at rest?

First, we need to determine the total rest energy of the final particles in the reaction. The rest energy of the proton is given as \(938\,\text{MeV}\), the neutral kaon as \(498\,\text{MeV}\), and the sigma baryon as \(1197\,\text{MeV}\). The total rest energy of these three particles is $$E_\text{rest-total} = 938 + 498 + 1197 = 2633\,\text{MeV}.$$ Now, we know that the electron and positron each have a rest energy of \(0.511\,\text{MeV}\). The minimum kinetic energy that they need to have is such that the total energy (rest energy + kinetic energy) of the electron-positron pair equals to the total rest energy of the final particles in the reaction. Let \(E_\text{k}\) be the kinetic energy of each of the electron and positron. We can write the conservation of energy equation as $$2 (0.511 + E_\text{k}) = 2633.$$ Next, solve for \(E_\text{k}\): $$E_\text{k} = \frac{2633 - 2\times 0.511}{2} \approx 1316\,\text{MeV}.$$ The minimum kinetic energy that each the electron and positron must have in order for the reaction to take place is approximately \(1316\,\text{MeV}\).

Since the sigma baryon decays at rest, its rest energy must be split between the rest energy and the kinetic energy of the decay particles, neutron and pion, during the decay process. We can obtain this through the conservation of energy equation: $$E_\text{rest-sigma} = E_\text{rest-neutron} + E_\text{rest-pion} + E_\text{k-neutron}+E_\text{k-pion}$$ Let \(E_\text{k-neutron}\) and \(E_\text{k-pion}\) be the kinetic energies of the neutron and pion, respectively. To find the kinetic energy of each decay particle, we can use the conservation of energy: $$1197\,\text{MeV} = 940\,\text{MeV} + 140\,\text{MeV} + E_\text{k-neutron}+E_\text{k-pion}$$ We can rewrite the equation as $$117\,\text{MeV} = E_\text{k-neutron}+E_\text{k-pion}$$ However, given the information in the problem, we cannot determine separate kinetic energies for the neutron and pion. The only thing we can say is that the sum of their kinetic energies must equal to \(117\,\text{MeV}\).