Question

This is Problem 4.53 from John Taylor Classical Mechanics

(a) Consider an electron (charge -e and mass m) in a circular orbit of radius r around a fixed proton (charge +e). Remembering that the inward Coulomb force ke²/r² is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to -1/2 times its PE; that is, T = -1/2 U and hence E = 1/2 U. Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius r around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches from afar with kinetic energy T_2. When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius r'.

(b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.)

(c) Identify the values of all five terms and the total energy E long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables T2, r, and r'.

Answer 1