Physics 195bProblem set number 17 – Solution to Problem 81Due 2 PM, Thursday, February 27, 2003READING: Read the “Identical Particles” course note.PROBLEMS:80. High energy limit: Do Exercise 7 of the Scattering course note.81. Consider the graph in Fig. 1.050 100 2001500408012016001δδδπT (MeV)PhaseShift(degrees)Figure 1: Made-up graph of phase shifts δ0and δ1for elastic π+p scattering(neglecting spin).Assume that the other phase shifts are negligible (e.g.,“lowenergy”is reasonably accurate). The pion mass and energy here are suffi-ciently small that we can at least entertain the approximation of aninfinitely heavy proton at rest – we’ll assume this to be the case,in any event. Note that Tπis the relativistic kinetic energy of theπ+: Tπ=P2π+ m2π− mπ.47(a) Is the π+p force principally attractive or repulsive (as shown inthis figure)?Solution: The phase shifts are positive, indicating a dominantlyattractive potential.(b) Plot the total cross section in mb (millibarns) as a function ofenergy, from Tπ=40 to 200 MeV.Solution: The total cross section in terms of the partial wavephase shifts is:σT=4πk2∞ =0(2 +1)sin2δ (150)=4πk2(sin2δ0+3sin2δ1). (151)The kinetic energy Tπis related to k by Tπ=m2π+ k2− mπ,ork =T (T +2mπ). (152)To convert to millibarns, we multiply by:1 = (197 MeV-fm)210 mb/fm2=3.88 × 105MeV2mb. (153)(c) Plot the angular distribution of the scattered π+at energies of120, 140 and 160 MeV.Solution:dσdΩ= |12ik∞j=0(2j +1) e2iδj(k)− 1Pj(cos θ)|2(154)=14k2|e2iδ0(k)− 1+3(e2iδj(k)− 1) cos θ|2(155)=14k2[cos δ0− 1 + 3(cos δ1− 1) cos θ]2+[sinδ0+3sinδ1cos θ]2.(d) What is the mean free path of 140 MeV pions in a liquid hydrogentarget, with these “protons”?Solution: The cross section for 140 MeV pions is ∼ 260 mb. Thedensity of liquid hydrogen is 0.0708 g/cm3. The number densityis ρ == 4.2 × 1028m−3. The mean free path is thusλ =1σTρ=0.9m. (156)4882. Inelastic scattering: Do Exercise 8 of the Scattering course note.83. Exclusion principle and atomic states: Do Exercise 1 of the IdenticalParticles course