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## PowerPoint Slideshow about 'Absorbed Dose in Radioactive Media I' - Jims

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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

Introduction (cont.)

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

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 << 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

Case 2

- In a large radioactive object (i.e., with mean radius >> 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

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

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 < 0.1% in this case

Graph of the data in the table for a broad plane beam of 1-MeV -rays in water

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

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:

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

Illustration of the reciprocity theorem as applied to estimate the -ray dose at point P within a homogeneous, uniformly radioactive object V

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

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

Introduction (cont.)

- If ̅Rdvis 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

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

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

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

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 ̅A̅Fdv,V for n different -ray energy lines can be gotten by

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

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

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

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

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

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

Ratio of average absorbed doses in uniformly radioactive tissue spheres with/without surrounding non-radioactive tissue medium

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

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

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

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

Radioactive Disintegration Processes (cont.)

- In this connection it should be noted that, according to Einstein’s mass-energy equation E = mc2, the energy equivalent of rest mass is as follows:

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

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

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

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

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

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

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

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

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