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Introduction: Photon Interactions Con’t.
•
TERMA: (Total Energy Released per unit Mass)
The focus here is on the primary radiation beam or field’s energy, and its
attenuation. No mention at all is made of what happens with this beam energy that
is released. It is defined for photons as follows:
T≡
µ
Ψ
ρ
and has units
cm −1
J
J
[T ] =
⋅ 2 =
3
kg / cm cm
kg
It has the same units as dose, but is quite far from dose. Yet, it is the first step in
that direction...
•
KERMA: (Kinetic Energy Released per unit Mass)
A critical shift in focus now to how the energy gets deposited, but not yet dose
either. Kerma indicates how much of the released energy above gets to charged
particles. These charged particles may subsequently radiate photons again, but
that is not yet considered. It is defined as follows:
µ tr
K≡
Ψ
ρ
and has units
[K ] =
J
kg
It has the same units as dose, but is still not dose. Yet, it is the next step in that
direction...
BUT …
Lecture 4 MP 501 Kissick 2016
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•
We need to define the energy transfer coefficient, µtr,
and the mass energy transfer coefficient, µtr/ρ.
It is defined via the average energy transferred to charged particles per
interaction, T , by
•
µtr T µ
=
⋅
ρ hν ρ
→
µ
T
= tr
hν
µ
Critical to this definition is an understanding of T . For each, i, of n photon
interactions, there is an amount of energy given to charged particles: (ε tr ) i . Then,
n
T = (1 / n)∑ (ε tr ) i
i =1
•
NOTE: This average relates to µtr
For the Compton effect below (from Attix, page 135), the right side axis is T :
Note, we will use this figure for homework, etc. in this class:
See also Appendix D.1 in Attix !
hν’/hν
•
T /hν
This mass energy transfer coefficient is is also like a probability, the probability
that energy is transferred to charged particles.
Lecture 4 MP 501 Kissick 2016
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•
So, for a photon with energy, hν, interacting at a point, the energy transferred to
charged particles,
εtr, will depend on the interaction type with examples as follows:
hν
Auger
electron
Photoelectric
flouro
x-ray
−
If f Auger produced
electron
hν
ε tr = Te + TAuger
hν ’
ε tr = Te
Compton
electron
hν
positron
Pair Production
−
Then, T f or electron
needs to be f ound.
annihilation
γs
ε tr = Te + Te
−
nucleus
+
electron
Note for pair production (if not annihilation-in-flight):
ε tr = hν − 1.022MeV + (1.022MeV − 1.022MeV ) ,
i.e., annihilations leave right away …
•
Note that all one cares about here is the transfer at a point to charges, but
subsequent or other photons that carry some energy elsewhere, are not considered
here. Another type of Kerma coming soon … Eventually, what we really want is
energy absorbed “locally.”
•
Note also that
µtr µ
T
since
<
<1
ρ ρ
hν
Lecture 4 MP 501 Kissick 2016
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Collision KERMA: (can be equal to dose in some situations!) considers charged
particles subsequently radiating photons again.
A most critical shift in focus now happens to how the KINETIC energy gets
deposited, but not yet dose due to geometry concerns of the radiation transport.
The issue remaining, as we get close to dose, is locality: does what comes into a
point equal what leaves (handled more later). These charged particles may
subsequently radiate photons again, and that is now considered. It is defined as
follows:
Kc ≡
µ en
Ψ
ρ
and has units
[Kc ] =
J
kg
It has the same units as dose, and it is close to dose. It is the next to the last step
in that direction...
BUT … AGAIN …
•
We need to define the energy ABSORPTION coefficient, µen,
and the mass energy ABSORPTION coefficient, µen/ρ.
•
We also need to define the NET energy transferred to charged particles:
εtrnet
Specifically, deposited within the ranges of the charged particles. We want to stay
local, close to a “point.” However, we know that photons carry energy around and
those photons are not to be included here.
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Also, another definition is needed: The average fraction of incident kinetic
energy transferred by photon radiation, g:
g≡
average subsequent photon radiation
T
∴
µ en
µ
= (1 − g ) tr
ρ
ρ
Lecture 4 MP 501 Kissick 2016
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•
•
For low energy photons, and low z materials, g ~ 0.
So, the fraction of energy not radiatively transferred, and CAN THUS BE
DEPOSITED LOCALLY is (1-g). The local issue gets difficult at high energy
because the charged particle range is close to the photon mean free path!
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A measure of the amount of energy that stays local is thus:
n
(1 − g ) =
(1 / n)∑ (ε trnet ) i
i =1
n
(1 / n)∑ (ε tr ) i
i =1
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Let’s consider how to calculate ε trnet :
Now we have added bremsstralung photons, hν’’, to the figure we are used to now.
Note that these photons and the annihilation photons are “far” from the point of
interaction.
hν
hν ’
hν ’’
Compton
Te− = Teinitial
−
ε trnet = Te − hv' '
−
Te −final
final
Assuming: Te − = 0
electron
Teinitial
+
hν
positron
Pair Production
nucleus
Teinitial
−
final
Annihilation, if “in-flight” then Te+
γs
final
Electron: often, Te −
hν ’’
=0
≠0
ε trnet = [Teinitial − Te final ]+
[T
−
initial
e+
−
]
− Te +final − hν ' '
Lecture 4 MP 501 Kissick 2016
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If no annihilation in flight: ε tr = ε trnet = hν − 1.022MeV + (1.022MeV − 1.022MeV )
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The photoelectric effect dominates at low energy where the subsequent
bremsstrahlung is negligible for the photoelectron and certainly the Auger
electrons. Therefore, hν’’~ 0, and ε trnet = ε tr :
Photoelectric effect
flouro
x-ray, hν’
M
L
K
hν << MeV
Auger
electron,
TAuger
ε trnet = Te + TAuger
−
photoelectron, Te
{Rock always felt, as do I, that these issues are not covered well in Attix.}
The central issue should be getting clearer now:
to connect measured (dose) and calculated (collision Kerma) quantities:
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All of the above quantities (Terma, Kerma, collision Kerma) are point or continuous
calculated quantities as much as possible. They are all proportional to energy
fluence.
•
Dose, however, is measured quantity (fundamentally), and can be related to the
above in some circumstances! Much of this class will examine how to connect dose
to collision Kerma.
•
Dose is measured in a volume, it has to be. There is a concept called “equilibrium”
that allows for all points inside that volume to be the same in the relevant ways, but
that will come later. For now,
Lecture 4 MP 501 Kissick 2016
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Consider just the energy imparted,
ε -- the energy deposited in a specified
volume, v:
ε = hν − hν '− Teout
−
ε = hν −[Teout + Teout ]−1.022MeV
−
−
out
ε = hν − hν '− (Teout + TAuger
)
−
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The photoelectric effect dominates at low energy and here too it leads to a
simplification that is usually true for low energies: T out
& T out ~ 0
−
e
Auger
Lecture 4 MP 501 Kissick 2016
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The dose is now just the expectation value of the energy imparted divided by the
mass of the small* volume:
Dose ≡
∆ε
(∆m = ρ∆V )
*Ideally, the region v or ∆v should be small enough so that all photons escape.
It is not possible to write an equation relating the absorbed dose directly to
the fluence or energy fluence. Therefore, the volume size of v is ideally dv,
but is never that in practice. See Attix pages 26 and 27.
Lecture 4 MP 501 Kissick 2016
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Examples from Attix: pages 28 and 29:
1)
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2)
i.e.,
(hν 1 − 1.022MeV ) − T3 + (1.022 MeV − 1.022MeV ) = T1 + T2 − T3
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Before going further, I have not used a basic quantity in Attix: Radiant Energy , R,
and the various forms of it. It is just the energy contained in the radiation:
charged and uncharged. If the sum of all radiant energy sources (like radioactive
isotopes) and sinks is
∑ Q , then the energy imparted can be written as
ε = (Rin − Rout )charged + (Rin − Rout )uncharged + ∑ Q
Please see Attix for rigorous definitions, but note that this ε applies to both
charged and uncharged particles! Note that ε tr and ε trnet apply only to photons.
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We will return to this later, but the trick will be to equate: Dose ~ K c . Possible?
•
Before that, any reasonable dose measurement needs to consider the effects of
scatter and geometry: the volume v above necessitates the move away from a pure
point quantity. It also brings to mind the issue of the finiteness of geometry.
Therefore, let’s extend the narrow beam discussion to …
Lecture 4 MP 501 Kissick 2016
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Broad beam geometry of Photons: consider the following measurement geometry
with a loss of collimation:
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Ideally here, lateral in-scattering equals lateral out-scattering, and then
Rdet (l ) = R0 det e − µen ⋅l
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However, we are not ever in the ideal situation, and the relationship between inscattering and out-scattering can be complex. Mostly, though,
out-scattering > in-scattering
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In that case, the effective attenuation coefficient, µ’:
µen < µ ' < µ
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(out-scattering > in-scattering)
Sometimes, the effective attenuation coefficient is negative !
corresponds to
out-scattering < in-scattering.
That case
Lecture 4 MP 501 Kissick 2016
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Attenuation/scatter (sphere)
detector
Point source
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This situation tends to happen at shallow depths. See Attix figures (page 49):
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This all leads to the concept of the Build-up factor, B.
B is a measure of the beam intensity to the primary (i.e., photons that have not yet
interacted) beam component intensity:
Lecture 4 MP 501 Kissick 2016
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B≡
•
intensity of primary + secondary radiation
intensity of primary radiation only
IF there is no inverse-square law (not point source, parallel beam instead), then:
R (l ) = R0 B (l )e − µl = R0 e − µ '( l ) l
Such that the effective attenuation coefficient is:
µ ' (l ) = µ −
ln B(l )
l
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In a narrow beam geometry, B=1, but in general, B>1. Also note the function of
distance or depth, l.
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Cautionary Note: Robert Jeraj’s colleague in Slovenia, Tomaz Podobnik has explored
Chapter 3 in Attix for issues surrounding Eq. 3.16. He suggests, and I think he is
correct, that one should correct Chapter 3 as follows to make the whole discussion
more consistent:
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Use Eq. 3.14 to define B(L) and write Eq. 3.15 as just (no middle term):
ψL
= e − µ 'L
ψ0
-- Then, Eq. 3.16 will follow more rigorously.
-- With µ ' ( L) defined as the slope of the broad-beam attenuation curve to give the
following:
Lecture 4 MP 501 Kissick 2016
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µ ' ( L) =
1 L
µ ' ( x)dx − ln B0 
∫


0

L
-- Note that it is not an average over energy even though the same ‘bar’ is used for
energy averages elsewhere in the chapter. Note in Figure 3.7 that µ ' points to
the average slope.
-- The backscatter factor, B0 , should be unity to make µ ' an average of µ ' over
the distance L.
-- However, a general expression would have a B0 > 1 , but that would diverge in the
limit of L approaches 0.
-- Therefore, the two lines at the bottom of page 54 are not correct.
-- In addition, in Figure 3.4(b), the small distance limit implies B0 > 1 , but the
guiding line extrapolates to what is effectively B0 = 1 , so the data on this graph
could have been better extrapolated with a better guiding line that does not
intersect Q / Q0 = 1 at L=0.
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We will go into details on all of this later, but here is a summary of the dose
related quantities for the main photon interactions:
Lecture 4 MP 501 Kissick 2016
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Lecture 4 MP 501 Kissick 2016