An  particle with initial kinetic energy K = 800 MeV

1. An  particle with initial kinetic energy K = 800 MeV moves head on toward a gold (Z=29) nucleus. It is slowed and then momentarily stopped at a distance of closest approach r0 from the nuclear center before being reversed (scattered by an angle  = 180o). Find r0. Because of the large energies involved we can assume the closest approach will not be affected by the presence of any atomic electrons. The repulsive Coulomb force changes K → PE 279 800 MeV = 1.44 MeV-fm = 28.4 fm The gold radius 6.4 fm so the  does not penetrate the gold nucleus.

2. Recall the cross section is defined by the ratio rate particles are scattered out of beam rate of particles focused onto target material/unit area a “counting” experiment notice it yields a measure, in units of area number of scattered particles/sec incident particles/(unit area  sec)  target site density how tightly focused or intense the beam is density of nuclear targets With a detector fixed to record data from a particular location ,  we measure the “differential” cross section: d/d.

3. 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 -

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

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

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

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

8. that general description of cross section we augmented with the specific example of Coulomb scattering

9. 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

11.  When >0 The incoming projectile will not reach r0, but a distance of closest approach rmin>r0 At that position which of course: Conservation of angular momentum guarantees:

12. impact parameter, b

13.  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:

14.  b q2 d and so

15.  b q2 d