Absorbed Dose in Radioactive Media I

1
Absorbed Dose in Radioactive Media I

• Introduction
• Alpha Disintegration

2
Introduction

• In this section we will consider radioactive
  processes and the deposition of absorbed dose in
  radioactive media
• Computation of the absorbed dose is
  straightforward for either CPE or RE conditions,
  but is more difficult for intermediate situations
• If the radiation emitted consists of charged
  particles plus much longer-range ?-rays, as is
  often the case, one can determine if CPE or RE is
  present, depending on the size of the radioactive
  object

3
Introduction (cont.)

• Assuming the conditions for RE are satisfied, and
  referring to the following diagram, we can
  consider two limiting cases

4
Radiation equilibrium
5
Case 1

• In a small radioactive object V (i.e., having a
  mean radius not much greater than the maximum
  charged-particle range d), CPE is well
  approximated at any internal point P that is at
  least a distance d from the boundary of V
• If d ltlt 1/? for the ?-rays, the absorbed dose D
  at P approximately equals the energy per unit
  mass of medium that is given to the charged
  particles in radioactive decay (less their
  radiative losses), since the photons practically
  all escape from the object and are assumed not to
  be scattered back by its surroundings

6
Case 2

• In a large radioactive object (i.e., with mean
  radius gtgt 1/? for the most penetrating ?-rays),
  RE is well approximated at any internal point P
  that is far enough from the boundary of V so
  ?-ray penetration through that distance is
  negligible
• The dose at P will then equal the sum of the
  energy per unit mass of medium that is given to
  charged particles plus ?-rays in radioactive decay

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Introduction (cont.)

• Deciding upon a maximum ?-ray range for case 2
  requires some kind of quantitative criterion
• Less than 1 of primary ?-rays penetrate through
  a layer 5 mean free paths in thickness, and less
  than 0.1 through 7 mean free paths
• However, one must take at least crude account of
  the propagation of the scattered photons, since
  we are dealing with a type of broad-beam geometry

8
Introduction (cont.)

• Referring to the example in the following diagram
  (1-MeV ?-rays, broad plane beam in water), we see
  that at a depth of 7 mean free paths the true
  attenuation is closer to 10-2 than 10-3
• Roughly 10 mean free paths are evidently
  necessary to reduce beam penetration to lt 0.1 in
  this case

9
Graph of the data in the table for a broad plane
beam of 1-MeV ?-rays in water
10
Introduction (cont.)

• Assuming the straight-ahead approximation
  (i.e., substituting ?en for ?? as an effective
  attenuation coefficient) would evidently require
  about 16 mean free paths (L ? 16/? ? 7/?en) to
  reach 0.1 penetration
• One concludes that if the relevant data for
  buildup factors or effective attenuation
  coefficients ?? are not available for a
  particular situation, the use of the
  straight-ahead approximation will overestimate
  the size of a radioactive object necessary to
  approximate RE at its center within desired
  limits
• Assuming ?? ? will underestimate that size by
  ignoring scattered rays

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Introduction (cont.)

• To estimate the ?-ray dose at an internal point
  in an intermediate-sized radioactive object, it
  will be helpful to define a quantity called the
  absorbed fraction

12
Introduction (cont.)

• The following figure illustrates the situation to
  be considered
• The volume V, representing the radioactive
  object, is filled by a homogeneous medium and a
  uniformly distributed ?-ray source
• It may be surrounded either by (1) an infinite
  homogeneous medium identical to that in V, but
  nonradioactive, or (2) an infinite vacuum

13
Illustration of the reciprocity theorem as
applied to estimate the ?-ray dose at point P
within a homogeneous, uniformly radioactive
object V
14
Introduction (cont.)

• In the first case a ?-ray escaping from V may be
  scattered back in in the second case it will be
  irrevocably lost
• The first case simulates more closely an organ in
  the body the second an object surrounded by air

15
Introduction (cont.)

• In case (1), the energy spent in the volume dv,
  at an internal point of interest P, by ?-rays
  from the source in dv at any other internal
  point P, is equal to the energy spend in dv by
  the source in dv
• The reciprocity theorem is exact in this case,
  because of the infinite homogeneous medium
• Since this equality holds for all points P
  throughout V, we may conclude that the energy
  spent in dv by the source throughout V is equal
  to the energy spent throughout V by the source in
  dv

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Introduction (cont.)

• If Rdv is the expectation value of the ?-ray
  radiant energy emitted by the source in dv and
  ?dv,V the part of that energy that is spent in
  V, then the absorbed fraction with respect to
  source dv and target V is
• For very small radioactive objects (V ? dv) this
  absorbed fraction approaches zero for an
  infinite radioactive medium it equals unity

17
Introduction (cont.)

• It has already been shown that ?dv,V
  ?V,dv,where ?V,dv is the energy spent in dv by
  gamma rays from the source throughout V
• Thus we can make a substitution, obtaining

18
Introduction (cont.)

• If one can calculate the absorbed dose fraction,
  its value is equal to the ratio of photon energy
  spent in dv by the source throughout V to the
  energy emitted by the source in dv
• This equals the ratio of the photon absorbed dose
  at P to that under RE conditions
• Thus if, say, 10 of the ?-ray energy from the
  source in dv escapes from V, this results in a
  10 reduction in ?-ray dose at P below its RE
  value, and AFdv,V 0.90

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Introduction (cont.)

• Assuming ?? to be the mean effective attenuation
  coefficient for ?-ray energy fluence transmission
  through a distance r of the medium, the fraction
  escaping from V in the direction of r from point
  P is e-??r
• In terms of the polar coordinates, with point P
  at the origin, the value of the absorbed dose
  fraction is given by

20
Introduction (cont.)

• Carrying out this integration is complicated by
  the fact that ?? (or the buildup factor B) is a
  function of r as well as h? moreover some
  radionuclides emit ?-rays of many different
  energies
• An average value of AFdv,V for n different
  ?-ray energy lines can be gotten by

21
Introduction (cont.)

• Calculation by Monte Carlo or moments methods
  have been reported by various authors
• A simple example of these results is shown in the
  following figure, which gives the radius of a
  unit-density tissue sphere required to produce
  absorbed fractions of 0.5 and 0.9, as a function
  of the photon quantum energy of a point source at
  the center

22
Radius of unit-density tissue sphere needed to
absorb 50 and 90 of the emitted photon energy
from a central point source in an infinite
homogeneous medium
23
Introduction (cont.)

• This example applies to case (1), in which the
  sphere is part of an infinite homogeneous medium
• The figure also gives the ?-dose at the center of
  the sphere if it were uniformly radioactive, as a
  fraction of the RE ?-dose there

24
Introduction (cont.)

• The dose decreases gradually as the point of
  interest is moved from the center of a ?-active
  object towards its boundaries
• In a radioactive volume V large enough to have RE
  at its center, imbedded in an infinite
  homogeneous medium, the ?-ray dose is reduced by
  approximately half in moving to the boundary of V
• At the interface between a semi-infinite
  homogeneous radioactive medium and another
  semi-infinite volume of the same medium without
  radioactivity, the dose would of course be
  exactly ½ of its RE value

25
Introduction (cont.)

• For purposes of internal dosimetry, one may be
  interested in the average ?-ray dose within a
  radioactive organ, rather than the dose at some
  specific point
• For this purpose one wants the value of AFV,V,
  which is just the average of AFdv,V for all
  points P throughout the volume V
• The Medical Internal Radiation Dose (MIRD)
  literature is mostly focused on such average-dose
  calculations

26
Introduction (cont.)

• Case (2), for which the volume V is surrounded by
  a void, is more difficult to calculate
• The reciprocity theorem is only approximate in
  that case because of the lack of backscattering
• Ellet has calculated the average absorbed dose in
  a uniformly radioactive tissue sphere of 780 g
  (5.7-cm radius) and 2.3 kg (8.2-cm radius), with
  and without a surrounding tissue scattering
  medium
• The results are given in the following diagram

27
Ratio of average absorbed doses in uniformly
radioactive tissue spheres with/without
surrounding non-radioactive tissue medium
28
Introduction (cont.)

• To obtain a crude estimate of the dose at some
  point P within a uniformly ?-active homogeneous
  object, it may suffice to obtain the average
  distance r from the point to the surface of the
  object, either by inspection or by using

29
Introduction (cont.)

• Then one may employ ?en ?? in the
  straight-ahead approximation to obtain
• which roughly approximates the ratio of ?-ray
  dose at P to that present if RE conditions existed

30
Radioactive Disintegration Processes

• Radioactive nuclei, either natural or
  artificially produced by nuclear reactions, are
  unstable and tend to seek more stable
  configurations through expulsion of energetic
  particles, including one or more of the
  following, where corresponding changes in the
  atomic number (Z) and number of nucleons (A) are
  indicated

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Radioactive Disintegration Processes (cont.)
32
Radioactive Disintegration Processes (cont.)

• The total energy (mass, quantum, and kinetic) of
  the photons and other particles released by the
  disintegration process is equal to the net
  decrease in the rest mass of the neutral atom,
  from parent to daughter
• Energy, momentum, and electric charge are each
  conserved in the process

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Radioactive Disintegration Processes (cont.)

• In this connection it should be noted that,
  according to Einsteins mass-energy equation E
  mc2, the energy equivalent of rest mass is as
  follows

34
Alpha Disintegration

• Alpha disintegration occurs mainly in heavy
  nuclei
• An important example is the decay of radium to
  radon, represented by the following mass-energy
  balance equation
• where ?½ symbolizes the half-life, or the
  time needed for ½ of the original number of
  parent atoms of Ra-226 to decay to the
  daughter product, Rn-222

35
Alpha Disintegration (cont.)

• Each of the elemental terms in this equation (and
  in other mass-energy equations to follow)
  represents the rest mass of a neutral atom of
  that element
• Notice that when the ?-particle (He nucleus) is
  emitted by the Ra-226 atom, its atomic number
  decreases by 2 and it consequently sheds two
  atomic electrons from its outermost shell, to
  become a neutral atom of Rn-222
• After the ?-particle slows down it captures two
  electrons from its surroundings, thereby becoming
  a neutral He atom

36
Alpha Disintegration (cont.)

• The 4.78 MeV shown in the equation is the energy
  equivalent of the rest mass decrease in
  transforming a neutral Ra-226 atom into neutral
  atoms of Rn-222 He-4
• It nearly all appears as particle kinetic energy,
  except for a small part that is given to 0.18-MeV
  photons, as discussed below

37
Alpha Disintegration (cont.)

• The corresponding mass-energy-level diagram for
  this disintegration is shown in the following
  diagram, where the vertical scale is given in
  terms of relative values of neutral atomic masses
  or their energy equivalents, as it will in later
  diagrams for other types of disintegrations

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39
Alpha Disintegration (cont.)

• Two branches are available for the disintegration
  of Ra-226
• 94.6 of these nuclei decay directly to Rn-222,
  making available 4.78 MeV, which is shared as
  kinetic energy between the ?-particle (4.70 MeV)
  and the recoiling Rn-222 atom (85 keV), the
  shares being proportional to the reciprocal of
  their masses to conserve momentum

40
Alpha Disintegration (cont.)

• The alternative branch for the decay of Ra-226
  occurs in only 5.4 of the nuclei, which release
  4.60 MeV of kinetic energy and give rise to a
  nuclear excited state of Rn-222
• This promptly relaxes to the ground state through
  the emission of a 0.18-MeV ?-ray
• The same total kinetic quantum energy is
  released by either route, and the net reduction
  in atomic rest mass is identical for each

41
Absorbed Dose from?-Disintegration

• In computing the absorbed dose in a medium from
  radioactive disintegration processes the
  calculation is always made under the assumption
  of the nonstochastic limit, and therefore the
  average branching ratios are used
• Again considering our example of radium decaying
  to radon, the average kinetic energy given to
  charged particles per disintegration is equal to

42
Absorbed Dose from?-Disintegration

• Under CPE conditions in a small (1-cm-radius)
  radium-activated object, if n such
  disintegrations occurred in each gram of the
  matter, the resulting absorbed dose would be
  given by 4.77n MeV/g, convertible into more
  conventional dose units
• Under RE conditions, on the other hand, the dose
  for the same concentration of radium would be
  simply 4.78n MeV/g, since the additional ?-ray
  energy would then be included