Protons of kinetic energy (in the centre-of-mass frame

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Gravitation and central-force motion · Gravitation and central-force motion

Action and Symmetry, Ch Scattering in a central field

Problem

Protons of kinetic energy EE (in the centre-of-mass frame, with the target nucleus treated as fixed because it is far heavier) are scattered by the Coulomb field of a nucleus of charge ZeZ\mathrm{e}. The deflection function is tan⁡(χ/2)=k/(2Eb)\tan(\chi/2) = k/(2Eb), with k=Ze2/4πε0k = Z\mathrm{e}^2/4\pi\varepsilon_0, bb the impact parameter and χ\chi the centre-of-mass scattering angle. ((a)) Starting from conservation of the number of particles passing per second through an annulus of the beam between impact parameters bb and b+dbb+\mathrm{d}b, and arriving in the cone of solid angle between χ\chi and χ+dχ\chi+\mathrm{d}\chi, derive dσ/dΩ=(b/sin⁡χ) ∣db/dχ∣\mathrm{d}\sigma/\mathrm{d}\Omega = (b/\sin\chi)\,|\mathrm{d}b/\mathrm{d}\chi|. ((b)) Use the stated Coulomb deflection function to eliminate bb and db/dχ\mathrm{d}b/\mathrm{d}\chi from this expression and show that dσ/dΩ=(k/4E)2/sin⁡4(χ/2)\mathrm{d}\sigma/\mathrm{d}\Omega = (k/4E)^2/\sin^4(\chi/2). ((c)) Evaluate dσ/dΩ\mathrm{d}\sigma/\mathrm{d}\Omega at χ=90∘\chi = 90^\circ for 3.0 MeV3.0\ \mathrm{MeV} protons scattered by gold (Z=79Z=79), taking e2/4πε0=1.44 MeV fm\mathrm{e}^2/4\pi\varepsilon_0 = 1.44\ \mathrm{MeV\,fm}.

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