1. Interaction of Gamma-Rays - General Considerations uncharged transfer of energy creation of fast electrons

2. photoelectric Compton pair production photodisintegration Interaction of Gamma-Rays - Types

3. Photoelectric Effect Einstein, 1905, as part of his Nobel prize winning paper on the photon theory of light - a prediction which was later verified experimentally in detail photon absorbed by atom, goes into excited state and ejects an electron with excess kinetic energy:

6. Photoelectric Effect

7. but  can also be expressed as a work function, a constant term "eW o " which varies from material to material  eV S = h - eW o where: h (the slope) remains constant for all material being equal to Planck's constant; 6.6  10 -34 J-s

8. there exists a threshold frequency: hV th = eW o below this threshold photons will not have sufficient energy to release even the least tightly bound electrons cross-section drops abruptly at the K-edge because below there is insufficient energy to overcome the binding and the K-shell electrons no longer participate the L-edge is really 3 energies due to fine level splitting Photoelectric Effect

9. K-shell binding energies  vary from 13.6 eV (H); 7.11 keV (Fe); 88 keV (Pb); 116 keV (U); if hν < E k, only L and higher shell electrons can take part photoelectric effect favored by low-energy photons and high Z absorbers; cross-section varies as: strong Z-dependence makes Pb a good x-ray absorber, usually followed by Cu and Al in a layered shield Photoelectric Effect

10. as vacancy left by the photoelectron is filled by an electron from an outer shell, either fluorescence x-rays or Auger electrons may be emitted the probability of x-ray emission is given by the "fluorescent yield" Photoelectric Effect

11. wave interpretation predicts that when electromagnetic radiation is scattered from a charged particle, the scattered radiation will have the same frequency as the incident radiation in all directions the scattering of electromagnetic radiation from a charged particle is viewed as a perfectly elastic billiard ball Compton Scattering

12. 4 unknowns: E 1, , K,  3 equations: momentum conservation (2), energy conservation γ Compton Scattering

13. K = (m - m o )c 2  difference between the total energy E of the moving particle and the rest energy E o (at rest) must treat electron relativistically Compton Scattering

14. while for photons energy K =(p 0 – p 1 )c Compton Scattering

15. momentum: Compton Scattering

16. substitute these expressions for K 1, p 2 into relativistic electron energy expression to get (after manipulation): where Δ is the shift that the scattered photon undergoes the wavelength is usually measured in multiples of "Compton units", the ratio of: Compton Scattering

17. wavelength the difference in energy E o - E 1 = K is the kinetic energy of the electron Compton Scattering

18. the min. electron energy corresponds to min. scattered photon energy (θ = 180º) so that this energy K max is called the Compton edge another form for this equation which uses photon energies instead of wavelengths is: Compton Scattering

19. at high incident energies E o the back scattered photon approaches a constant energy Compton Scattering

20.  0.511 MeV θ = 90º  0.255 MeV θ = 180º in this limit we find that 0 «  ≈, so that the energy: Compton Scattering

21. In a Compton experiment an electron attains kinetic energy of 0.100 MeV when an x-ray of energy 0.500 MeV strikes it. Determine the wavelength of the scattered photon if the electron is initially at rest Compton Scattering Problem

22. in the process of pair production the energy carried by a photon is completely converted into matter, resulting in the creation of an electron-positron pair σ pp ~ Z 2 since the charge of the system was initially zero, 2 oppositely charged particles must be produced in order to conserve charge Pair Production

23. in order to produce a pair, the incident photon must have an energy of the pair; any excess energy of the photon appears as kinetic energy of the particles Pair Production

25. pair production cannot occur in empty space the nucleus carries away an appreciable amount of the incident photon's momentum, but because of its large mass, its recoil kinetic energy, k ≈ p 2 /2m o, is usually negligible compared to kinetic energies of the electron-positron pair thus, energy (but not momentum) conservation may be applied with the heavy nucleus ignored, yielding: h = m + c 2 + m - c 2 = k + + k - + 2m o c 2 since the positron and the electron have the same rest mass; m o = 9.11 x 10 -31 kg Pair Production

26. the inverse of pair production can also occur in pair annihilation a positron-electron pair is annihilated, resulting in the creation of 2 (or more) photons as shown at least 2 photons must be produced in order to conserve energy and momentum Annihilation

27. in contrast to pair production, pair annihilation can take place in empty space and both energy and momentum principles are applicable, so that:  where k l is the propagation vector: 2 (k) = 2  / Annihilation

30. problem: how many positrons can a 200 MeV photon produce? the energy needed to create an electron-positron pair at rest is twice the rest energy of an electron, or 1.022 MeV; therefore maximum number of positrons equals: Annihilation

31. absorber nucleus captures a  -ray and in most instances emits a neutron: 9 Be( ,n) 8 Be important for high energy photons from electron accelerators cross-sections are « total cross-sections Photodisintegration

32. total attenuation coefficient  in computing shielding design the above equation is used this is the fraction of the energy in a beam that is removed per unit distance of absorber Combined Effects

33. the fraction of the beam's energy that is deposited in the absorber considers only the energy transferred by the photoelectron, Compton electron, and the electron pair energy carried away by the scattered photon by Compton and by annihilation is not included  Combined Effects

34. Linear Attenuation and Absorption Coefficients for Photons in Water

36. Exponential Absorption due to the different interaction of  -rays with matter, the attenuation is different than with α or  particles intensity of a beam of photons will be reduced as it passes through material because they will be removed or scattered by some combination of photoelectric effect, Compton scattering and pair production reduction obeys the exponential attenuation law:

37. I = I o e -  t where: I 0 =  -ray intensity at zero absorber thickness t = absorber thickness I =  -ray intensity transmitted  = attenuation coefficient if the absorber thickness is measure in cm, then  is called linear attenuation coefficient (  l ) having dimensions "per cm" Exponential Absorption

38. if the absorber thickness "t" is measured in g/cm 2, then (  m ) is called mass attenuation coefficient (  m ) having dimensions cm 2 /g where  is the density of the absorber Exponential Absorption

39. what percent of an incident x-ray passes through a 5 mm material whose linear absorption is 0.07 mm-1?

40. a monochromatic beam of photons is incident on an absorbing material if the intensity is reduced by a factor of 2 by 8 mm of material, what is the absorption coefficient? Exponential Absorption

41. Half-Value Thickness (HVT) thickness of absorber which reduces the intensity of a photon beam to 1/2 its incident value find HVT of aluminum if  = 0.070 mm -1 Exponential Absorption

42. Atomic Attenuation Coefficient  a fraction of an incident  -ray beam that is attenuated by a single atom, or the probability that an absorber atom will interact with one of the photons where  a is referred to as a cross-section and has the units barns Exponential Absorption

43. Linear Attenuation Coefficients

44. what is the thickness of Al and Pb to transmit 10% of a 0.1 MeV  -ray? (0.435 cm -1 ) Exponential Absorption

45. if we have a 1.0 MeV  - ray: compute the density thickness at 0.1 MeV Exponential Absorption

46. at 1.0 MeV this shows that Pb is only slightly better on a mass basis than Al however for low energy photons Pb is much better in general, for energies between 0.8  5 MeV almost all materials, on a mass basis, have approximately the same  -ray attenuating properties Exponential Absorption

49. 1-MeV photons are normally incident on a 1- cm lead slab the mass attenuation coefficient of lead (density = 11.35 g/cm 3 ) is 0.0708 cm 2 /g and the atomic weight is 207.2 Photon Interactions - Problem

50. calculate the linear attenuation coefficient what fraction of 1-MeV photons interact in a 1-cm lead slab? what thickness of lead is required for half the incident photons to interact? calculate the mean free path Photon Interactions - Problem

51. Solution a. the linear attenuation coefficient  is obtained by multiplying the mass attenuation coefficient by the density  = (0.0708 cm 2 g -1 )  (11.35 g cm -3 = 0.804 cm -1. b. if I o photons are incident on the lead slab and I photons penetrate it without interacting, then the fraction not interacting is given by: I/I o = exp[-  x] = exp[(-0.804 cm -1 )(1 cm)] = 0.448 Photon Interactions - Problem the fraction of photons interacting is then:  1 - 0.448 = 0.552

52. c.substituting I/I o = 1/2 in eq. (1) above and rearranging yields: x = ln 2 /μ = 0.693/0.804 = 0.862 cm Photon Interactions - Problem

53. d.the mean free path (MFP) is the average distance that an incident photon travels before interaction and is the reciprocal of the attenuation coefficient: MFP = 1/μ = 1/0.804 = 1.24 cm this is also numerically equal to the "relaxation length“ the relaxation length is the shield thickness needed to attenuate a narrow beam of monoenergetic photons to 1/e (= 0.368) of its original intensity and can be derived by substituting I/I o = 1/e Photon Interactions - Problem