1. Imagine a narrow, well-collimated beam of mono-energetic particles passing through a slab of matter EoEo E EoEo

2. EoEo E Energy loss E1E1 EE

3. 0 beam direction,   Similarly, the initially well-defined beam direction suffers at least small-angle scattering:

4. EoEo E Energy loss E1E1 EE If the target is thick, this implies that the overall mean energy loss  thickness For sufficiently high initial E 0 (or thin enough targets) all particles get through.

5. EoEo E 0 but if the target is thick enough thickness 

7. Besides simple scattering there are the (rarer) “all-or-nothing” interactions where the interaction ELIMINATES particles from the beam loss of beam particles for thin targets  thickness where 1/   the “mean free path”

8. EarthMoon

9. EarthMoon

10. In a solid interatomic spacing: 1  5 Å (1  5  10 -10 m) nuclear radii: 1.5  5fm (1.5  5  10 -15 m) for some sense of spacing consider the ratio orbital diameters central body diameter ~ 10s for moons/planets ~100s for planets orbiting sun the ratio orbital diameters central body diameter ~ 66,666 for atomic electron orbitals to their own nucleus A basketball scale nucleus would have its family of electrons stretching 10s of miles away

11. Carbon 6 C Oxygen 8 O Aluminum 13 Al Iron 26 Fe Copper 29 Cu Lead 82 Pb What about a single, high energy, charged particle? While the mass of matter is due primarily to it’s nuclei The volume of matter is due primarily to it’s electron clouds

12. A solid sheet of lead offers how much of a (cross sectional) physical target (and how much empty space) to a subatomic projectile? 82 Pb 207 Number density, n : number of individual atoms (or scattering centers!) per unit volume w n= (11.3 g/cc)(6.02  10 23 /mole)/(207.2 g/mole) = 3.28  10 22 /cm 3 n=  N A / A where N A = Avogadro’s Number A = atomic weight (g)  = density (g/cc)

13. What fraction of the target area physically has something in a sub-atomic projectile’s way? For a thin enough layer n  ( Volume )  ( atomic cross section ) = n  (surface area  w)(  r 2 ) as a fraction of the target’s area: = n  (w)   13 cm) 2  -15 m For 1 mm sheet of lead:0.00257 1 cm sheet of lead: 0.0257

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

16. q2q2 Recoil of target BOTH target and projectile move in response to the forces between them. q1q1 q1q1    A particle of charge q 1 encountering (passing by, but not directly hitting) a heavy charge q 2 at rest, follows a HYPERBOLIC TRAJECTORY

17. q2q2 q1q1 b F F'F' For an attractive “central” force the heavy charge occupies the focus of the trajectory like the sun does for a comet sweeping past the sun (falling from and escaping back to distant space).

19. impact parameter, b 

20. dd q2q2  b 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  using becomes: d  = 2  b db

21. dd q2q2  b and from bottom of previous page so Starting back at:

22. dd q2q2  b

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

25. V V

26. V ??

27. mv + M(0) = mv mv = mv 1 + Mv 2 Conservation of Momentum: The initial momentum Must equal the final momentum ½mv 2 + ½M(0) 2 = ½mv 2 ½mv 2 = ½mv 1 2 + ½Mv 2 2 Conservation of Energy: The initial energy must equal the final energy

28. mv = mv 1 + Mv 2 ½mv 2 = ½mv 1 2 + ½Mv 2 2 both equations must be satisfied together mv  mv 1 = Mv 2 m(v  v 1 ) = Mv 2 2  ½mv 2 = 2  (½mv 1 2 + ½Mv 2 2 ) mv 2 = mv 1 2 + Mv 2 2 m(v 2  v 1 2 ) = Mv 2 2

29. m(v  v 1 ) = Mv 2 m(v 2  v 1 2 ) = Mv 2 2 m(v  v 1 ) = Mv 2 m(v 2  v 1 2 ) = Mv 2 2 Divide the top equation by the bottom equation (v  v 1 ) = v 2 (v 2  v 1 2 ) = v 2 2 (v  v 1 ) (v  v 1 ) = v 2 2 (v  v 1 ) = v 2

30. m(v  v 1 ) = Mv 2 m(v 2  v 1 2 ) = Mv 2 2 (v  v 1 ) = v 2 m(v  v 1 ) = M(v + v 1 ) mv  mv 1 = Mv + Mv 1 mv  Mv = mv 1 + Mv 1 (m  M)v = (m + M)v 1 (m  M) (m + M) v = v 1

31. (m  M) (m + M) v 1 = v 2m (m + M) v 2 = v If M = m:v 1 = v 2 = If M = 10m :v 1 = v 2 = If M = 100m: v 1 = v 2 = If M>>100m: v 1  0v0v -(9/11)v = -0.8182v (2/11)v = 0.1818v -(99/101)v = -0.9802v (2/101)v = 0.1980v -v

32. V

33. V (M  m) (m + M) v 1 = v 2M (m + M) v 2 = v To lowest order, when M>>m v 1  v

34. Then for Let’s look at:

35.  mv 0 mv f mv 0 mv f  (mv) =  recoil momentum of target ( ) For small scattering (  ) large impact parameter b and/or large projectile speed v 0 v f  v o

36. mv 0 mv f pp  /2 Together with: Recognizing that all charges are simple multiples of the fundamental unit of the electron charge e, we write q 1 = Z 1 e q 2 = Z 2 e

37. Recalling that kinetic energy K = ½mv 2 = (mv) 2 /(2m) the transmitted kinetic energy (the energy lost in collision to the target) K = (  p) 2 /(2m target )

38. q1=Z1eq1=Z1e q2=Z2eq2=Z2e Z 2 ≡Atomic Number, the number of protons (or electrons)

39. For nuclear collisions: m target  2Z 2 m proton For collisions with atomic electrons: m target  m electron q 1 = e

40. Z 2 times as many of these occur! For nuclear collisions: m target  2Z 2 m proton For collisions with atomic electrons: m target  m electron q 1 = e

41. The energy loss due to collisions with electrons is GREATER by a factor of m proton = 0.000 000 000 000 000 000 000 000 00 1 6748 kg m electron = 0.000 000 000 000 000 000 000 000 000 000 9 kg

42. Notice this simple approximation shows that Why are  -particles “more ionizing” than  -particles?

43. energy loss speed

44. We express the energy loss as But in practice the thickness, x, alone is not important. So we define an effective or relative length that incorporates the target material’s density: Notice this x does not carry “normal” length units!

45. Need to average this over all possible values of b. Through any thickness, subsequent encounters are with progressively smaller values of E so need to integrate over E 0 to E f. The electrons involved were assumed FREE. Only energy in excess of the ionization energy I participates in this sort of momentum exchange. A single encounter! Actually should be redone relativistically!

46. E (MeV) Range of dE/dx for proton through various materials Pb target H 2 gas target dE/dx ~ 1/  2 Logarithmic rise 10 3 10 2 10 1 10 0 10 1 10 2 10 4 10 5 10 6 -dE/dx = (4  N o z 2 e 4 /m e v 2 )(Z/A)[ ln {2m e v 2 /I(1-  2 )}-  2 ] I = mean excitation (ionization) potential of atoms in target ~ Z10 GeV Felix Bloch Hans Bethe

47. Particle Data Group, R.M. Barnett et al., Phys.Rev. D54 (1996) 1; Eur.Phys.J. C3 (1998) Muon momentum [GeV/c]  

48. D. R. Nygren, J. N. Marx, Physics Today 31 (1978) 46    p d e Momentum [GeV/c] dE/dx(keV/cm)

50. EXAM 1 Mean = 66.0 Standard deviation = 23.1  = 66.0 10 20 30 40 50 60 70 80 90 100 Hour exams ( 3 x 100) 300 Homework Problems 100 Final Exam 200

51. Electrons e  so light they can scatter madly, suffering large deflections & energy losses Ionization Region Radiation Region Two regions are defined by an emperical “Critical Energy” Z of target atoms dE dx just like protons, etc except range ( penetration depth ) is measured over actual path length Bremsstrahlung (braking radiation)

52. Bremsstrahlung braking radiation Photon energies radiated, E  Resulting in “all-or-nothing” depletion of the beam’s energy “radiation length” after ~7X 0 only 1/10 3 of the initial electron energy remains

53. 

54. Pair Production ee  ee E  = ? p  = ? h E/c=h /c from the relativistic energy expression: rest mass energy plus kinetic energy (total E=  mc 2 ) This process can only occur near other charged particles! Why?

55.  EE pp E1E1 E2E2 “opening angle” Can’t be created with just enough E=2mc 2

56. What do photos like this suggest? and think: in the COM frame that clearly MUST be true always positive! In the COM frame of the e + e  so but??? i.e., the assumed simple picture of  e + e  cannot be correct!

57. If Bremsstrahlung photons energetic enough > 1 MeV they can pair produce The mean free path of a (high energy) photon through matter is

58. e-e- e-e- e-e- e+e+  Bremsstrahlung Pair production Photon or “gamma” ray Counting interaction lengths

59. Number of n electrons in a shower as a function of the thickness traversed, t, in radiation lengths. B.Rossi and K.Greisen, Rev. of Modern Phys. 13, 240 (1941).

60. Photons interact within matter via 3 processes 1.The photo-electric effect 2.Compton scattering 3.Pair production  e-e- 1.

61. Compton Scattering   While the initial energy of the electron’s just p1p1 p2p2 meme Where: the photon’s electron energy binding energy >> so the electron is basically “free”

62.   p1p1 p2p2 meme Conservation of momentum Conservation of energy squaring :

63. equating both of these equations multiply each side by hc p 1 p 2 E o then recalling =h/p The change  is independent of ! o o

64. Total absorption coefficients of  rays by lead and aluminum as a function of energy. W. Heitler, The Quantum Theory of Radiation, The Clarendon Press, Oxford, 1936. ħ  /mc=1 corresponds to 511 keV E  ~keV Pair production impossible and Compton cross section dominates at E=2m e c 2 pair production turns on