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Interaction of Gamma-Rays - General ConsiderationsPowerPoint Presentation

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

- 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:

- 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

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

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

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"

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

Compton Scattering

γ

4 unknowns: E1,, K,

3 equations: momentum conservation (2),

energy conservation

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

Compton Scattering

- momentum:

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:

Compton Scattering

- wavelength

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

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:

Compton Scattering

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

Compton Scattering

0.511 MeV θ = 90º

0.255 MeV θ = 180º

- in this limit we find that 0« ≈ , so that the energy:

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

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

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

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

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

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/

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:

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

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

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

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:

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"

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

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?

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?

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

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

Linear Attenuation Coefficients in Water

(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?

Exponential Absorption in Water

- if we have a 1.0 MeV - ray:

- compute the density thickness at 0.1 MeV

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

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

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

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

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

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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