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v D t

scattered particles. Incident mono-energetic beam. v D t. A. d W. N = number density in beam (particles per unit volume). Solid angle d W represents detector counting the dN particles per unit time that scatter through q into d W. N number of scattering. centers in target

benjamin-stewart
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v D t

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  1. scattered particles Incident mono-energetic beam v Dt A dW N = number density in beam (particles per unit volume) Solid angle dWrepresents detector counting the dN particles per unit time that scatter through qinto dW Nnumber of scattering centers in target intercepted by beamspot FLUX = # of particles crossing through unit cross section per sec = NvDt A / Dt A = Nv Notice: qNv we call current, I, measured in Coulombs. dN NF dW dN = s(q)NF dW dN =NFds -

  2. dN = FNs(q)dWNFds(q) the “differential” cross section R  R R R R

  3. R R the differential solid angle d for integration is sin d d Rsind Rd Rsind Rd Rsin

  4. Symmetry arguments allow us to immediately integrate  out and consider rings defined by  alone R Rsind  R R R Nscattered= NFsTOTAL Integrated over all solid angles

  5. Nscattered= NFsTOTAL The scattering rate per unit time Particles IN (per unit time) = FArea(ofbeamspot) Particles scattered OUT (per unit time) = F NsTOTAL

  6. Earth Moon

  7. Earth Moon

  8. for some sense of spacing consider the ratio orbital diameters central body diameter ~ 10s for moons/planets ~100s for planets orbiting sun • In a solid • interatomic spacing:1-5 Å (1-5  10-10 m) • nuclear radii: 1.5 -5 f(1.5-5  10-15 m) the ratio orbital diameters central body diameter ~ 66,666 for atomic electron orbitals to their own nucleus Carbon 6C Oxygen 8O Aluminum 13Al Iron 26Fe Copper 29Cu Lead 82Pb

  9. A solid sheet of lead offers how much of a (cross sectional) physical target (and how much empty space) to a subatomic projectile? 82Pb207 w Number density,n: number of individual atoms (or scattering centers!) per unit volume n= rNA / A where NA = Avogadro’s Number A = atomic weight (g) r = density (g/cc) n= (11.3 g/cc)(6.021023/mole)/(207.2 g/mole) = 3.28  1022/cm3

  10. 82Pb207 w For a thin enough layer n(Volume)  (atomic cross section) = n(surface areaw)(pr2) as a fraction of the target’s area: = n(w)p(5  10-13cm)2 For 1 mm sheet of lead: 0.00257 1 cm sheet of lead: 0.0257

  11. Actually a projectile “sees” nw nuclei per unit area but Znw electrons per unit area!

  12. that general description of cross section let’s augmented with the specific example of Coulomb scattering

  13. BOTH target and projectile will move in response to the forces between them. q1   q2  Recoil of target But here we are interested only in the scattered projectile q1

  14. impact parameter, b

  15. b q2 d A beam of N incident particles strike a (thin foil) target. The beam spot (cross section of the beam) illuminates n scattering centers. If dN counts the average number of particles scattered between and d dN/N = n d d = 2 b db using becomes:

  16. b q2 d and so

  17. b q2 d

  18. What about the ENERGY LOST in the collision? • the recoiling target carries energy • some of the projectile’s energy was surrendered • if the target isheavy • the recoil is small • the energy loss is insignificant Reminder: 1/ (3672 Z)

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