Interaction of Gamma-Rays - General Considerations

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

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

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

8. Photoelectric Effect • there exists a threshold frequency: hVth = eWo • 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

9. Photoelectric Effect • K-shell binding energies  vary from 13.6 eV (H); 7.11 keV (Fe); 88 keV (Pb); 116 keV (U); if hν < Ek, 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

10. Photoelectric Effect • 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"

11. Compton Scattering • 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

12. Compton Scattering γ 4 unknowns: E1,, K,  3 equations: momentum conservation (2), energy conservation

13. Compton Scattering • K = (m - mo)c2 difference between the total energy E of the moving particle and the rest energy Eo (at rest) • must treat electron relativistically

14. Compton Scattering • while for photons • energy K =(p0 – p1)c

15. Compton Scattering • momentum:

16. Compton Scattering • substitute these expressions for K1, p2 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:

17. Compton Scattering • wavelength • the difference in energy Eo - E1 = K is the kinetic energy of the electron

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

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

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

21. Compton ScatteringProblem 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

22. Pair Production • 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 σpp ~ Z2 • since the charge of the system was initially zero, 2 oppositely charged particles must be produced in order to conserve charge

23. Pair Production • 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

24. Pair Production

25. Pair Production • 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 ≈ p2/2mo, 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+c2 + m-c2 = k+ + k- + 2moc2 • since the positron and the electron have the same rest mass; mo = 9.11 x 10-31 kg

26. Annihilation • 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

27. Annihilation • 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/

28. Annihilation

29. Annihilation

30. Annihilation 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:

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

32. Combined Effects • 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

33. Combined Effects • 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 

34. Linear Attenuation and Absorption Coefficients for Photons in Water

36. ExponentialAbsorption • 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. Exponential Absorption I = Ioe-t • where: I0 = -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"

38. Exponential Absorption • if the absorber thickness "t" is measured in g/cm2, then (m) is called mass attenuation coefficient (m) having dimensions cm2/g where  is the density of the absorber

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

40. Exponential Absorption • 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?

41. Exponential Absorption 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

42. Exponential Absorption 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

43. Linear Attenuation Coefficients

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

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

46. Exponential Absorption • 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

49. Photon Interactions - Problem • 1-MeV photons are normally incident on a 1-cm lead slab • the mass attenuation coefficient of lead (density = 11.35 g/cm3) is 0.0708 cm2/g and the atomic weight is 207.2

50. Photon Interactions - Problem • 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