Interaction of gamma rays general considerations
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Interaction of Gamma-Rays - General Considerations. uncharged transfer of energy creation of fast electrons. Interaction of Gamma-Rays - Types. photoelectric Compton pair production photodisintegration. Photoelectric Effect.

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Interaction of Gamma-Rays - General Considerations

  • uncharged

  • transfer of energy

  • creation of fast electrons


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Interaction of Gamma-Rays - Types

  • photoelectric

  • Compton

  • pair production

  • photodisintegration


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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:



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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


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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


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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


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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"


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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


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Compton Scattering

γ

4 unknowns: E1,, K, 

3 equations: momentum conservation (2),

energy conservation


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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


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Compton Scattering

  • while for photons

  • energy

K =(p0 – p1)c



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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:


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Compton Scattering

  • wavelength

  • the difference in energy Eo - E1 = K is the kinetic energy of the electron


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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:


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Compton Scattering

  • at high incident energies Eo the back scattered photon approaches a constant energy


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Compton Scattering

 0.511 MeV θ = 90º

 0.255 MeV θ = 180º

  • in this limit we find that 0«  ≈ , so that the energy:


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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


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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


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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



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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


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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


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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/




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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:


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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


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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


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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 



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Exponential in WaterAbsorption

  • 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:


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Exponential Absorption in Water

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"


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Exponential Absorption in Water

  • 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


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Exponential Absorption in Water

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


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Exponential Absorption in Water

  • 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?


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Exponential Absorption in Water

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


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Exponential Absorption in Water

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



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(0.435 cm in Water-1)

Exponential Absorption

  • what is the thickness of Al and Pb to transmit 10% of a 0.1 MeV -ray?


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Exponential Absorption in Water

  • if we have a 1.0 MeV  - ray:

  • compute the density thickness at 0.1 MeV


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Exponential Absorption in Water

  • 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


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Photon Interactions - Problem in Water

  • 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


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Photon Interactions - Problem in Water

  • 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


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Photon Interactions - Problem in Water

Solution

a. the linear attenuation coefficient  is obtained by multiplying the mass attenuation coefficient by the density

 = (0.0708 cm2 g-1)  (11.35 g cm-3 = 0.804 cm-1.

b. if Io photons are incident on the lead slab and I photons penetrate it without interacting, then the fraction not interacting is given by:

I/Io = exp[-  x]

= exp[(-0.804 cm-1)(1 cm)] = 0.448

the fraction of photons interacting is then: 

1 - 0.448 = 0.552


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Photon Interactions - Problem in Water

  • substituting I/Io = 1/2 in eq. (1) above and rearranging yields:

    x = ln2/μ = 0.693/0.804 = 0.862 cm


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Photon Interactions - Problem in Water

  • 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/Io = 1/e


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