Suppose an electron and a positron, each with rest evergies of 0.511 MeV, collid

Question

Suppose an electron and a positron, each with rest evergies of 0.511 MeV, collide to create a proton (rest energy 938 MeV), an electrically neutral kaon (498 MeV), and a negatively charged sigma baryon (1197 MeV), The reaction can be written as:

(a) What is the minimum kinetic energy the electron and positron must have to make this reaction go? Assume they have the same energy.

(b) The sigma can decay in the reaction to n+- with rest evergies of 940 MeV (neutron) and 140 MeV (pion). What is the kinetic energy of each decay particle if the sigma decays at rest?

Explanation / Answer

(a) If the particles after the decay are at rest, you can use conservation of energy to determine the amount of kinetic energy the electron and positron had together before the collision: 2*0.511MeV + KE = 938MeV + 498MeV + 1197MeV Solve for KE (b) Do the same here, but in reverse. The particle is at rest before the decay and the particles after have some KE: 1197MeV = 940MeV + 140MeV + KE_neutron + KE_pion, where KE_neutron and KE_pion are the kinetic energy of each particle, 0.5*m_neutron*(v_neutron)^2 and 0.5*m_pion*(v_pion)^2, respectively. You'll need an additional equation to solve for the velocities individually. Use conservation of momentum. 0 = m_neutron*v_neutron + m_pion*v_pion Now you have two equations and two unknowns, solve for one v in one equation and stick it into the other to solve for one velocity. Then plug that velocity back into one of the equations to find the other. Finally, what you really want is the kinetic energy of each particle, so stick both v's back into their respective equations for kinetic energy.

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