Alpha particles ( ) of centre-of-mass kinetic energy strike a silver foil (

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Mechanics & Dynamical Systems · Calculus of change and accumulation

Action and Symmetry, Ch Scattering in a central field

Problem

Alpha particles (Z1=2Z_1=2) of centre-of-mass kinetic energy E=6.0 MeVE=6.0\,\mathrm{MeV} strike a silver foil (Z2=47Z_2=47, molar mass 107.87 g mol−1107.87\,\mathrm{g\,mol^{-1}}, density 10.49×103 kg m−310.49\times10^3\,\mathrm{kg\,m^{-3}}) of thickness t=300 nmt=300\,\mathrm{nm}. Take k=Z1Z2×1.44 MeV fmk=Z_1Z_2\times1.44\,\mathrm{MeV\,fm} and Avogadro's number NA=6.022×1023 mol−1N_{\rm A}=6.022\times10^{23}\,\mathrm{mol^{-1}}. ((a)) Starting from the Rutherford differential cross-section dσ/dΩ=(k/4E)2/sin⁡4(χ/2)\mathrm{d}\sigma/\mathrm{d}\Omega=(k/4E)^2/\sin^4(\chi/2), derive an expression for the area per atom, σ(χ>χmin)\sigma(\chi>\chi_{\rm min}), that scatters an alpha beyond a stated angle χmin\chi_{\rm min}, by integrating over solid angle from χmin\chi_{\rm min} to π\pi. ((b)) Use this, together with the number of silver atoms per unit area in the foil, to find the fraction of the incident beam scattered beyond χmin=60∘\chi_{\rm min}=60^\circ.

Browse the problem library