(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 \(k e^{2} / r^{2}\) is what gives the electron its centripetal
acceleration, prove that the electron's KE is equal to \(-\frac{1}{2}\) times
its \(\mathrm{PE}\); that is, \(T=-\frac{1}{2} U\) and hence \(E=\frac{1}{2} U\).
(This result is a consequence of the so-called virial theorem. See Problem
4.41.) 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 \(r^{\prime} .\) (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 \(T_{2}, r,\) and \(r^{\prime}\).