Calculation of the Radiation Dose
To calculate the radiation dose, one has to determine the average amount of energy absorbed by 1 g of tissue of a target (organ of interest) from the total energy released by the decay of a given amount of radioactivity. Because x- or γ-rays are more penetrating than particulate radiation, a small x- or γ-ray emitting source, localized at any site in the body, irradiates practically every organ of the body. For example, an x- or γ-ray emitting radionuclide localized only in the liver (source S) delivers the radiation dose to all other organs (targets T) of the body in addition to the liver. Thus, if a radiopharmaceutical is localized in multiple organs (sources), the radiation dose from each source to each organ (target) has to be calculated before the final determination of the radiation dose to each organ can be made by adding all contributions. The following four steps are involved in the radiation-dose determinations:
Calculation of the rate of energy emission (erg/h) of the various types of radiation emitted by the radionuclide in the source volume.
Calculation of the rate of energy absorption from these radiations by the target volume.
Calculation of the average dose rate, dD/dt.
Calculation of the average dose D.
This method of radiation dose calculation is known as the absorbed fraction method. The first three steps require mainly physical data such as decay characteristics, organ shape and size, and so on, whereas the fourth step requires biologic distribution data.
Step 1—Rate of Energy Emission. Let us first consider a radionuclide that emits only one type of radiation (emission frequency = 1) of energy E (MeV) per decay. One microcurie (3.7 × 104 decay/s) of this radionuclide will, therefore, emit energy at a rate of 3.7 × 104 × E MeV/s. If we change the unit of energy from MeV to erg (1 MeV = 1.6 × 10-6 erg) and the unit of time from second to hour (1 h = 3600 s), the rate of energy emission by this radionuclide becomes equal to 3.7 × 104 × 1.6 × 10-6 × 3600 × E erg/(h × µCi), or 213E erg/(h × µCi).
In the case of a radionuclide that emits more than one radiation, say 1, 2, 3, … n, with emission frequencies n1, n2, n3, …, nn and energies of E1, E2, E3, …, En, respectively, the rate of energy emission for each type of radiation will be equal to 213n1E1 erg/(h × µCi) for radiation 1, 213n2E2 erg/(h × µCi) for radiation 2, and so on.
Step 2—Rate of Energy Absorption. To calculate the rate of absorption of energy by a target volume T from a radionuclidic distribution in a source volume S, we have to define a new quantity known as the absorbed fraction, φi(T←S). The absorbed fraction φi(T←S) is defined as the ratio of energy absorbed by a target volume T from a radiation i to the amount of energy released by a radionuclidic distribution in volume S in the form of radiation i. In other words
In most problems encountered in nuclear medicine, the radioactivity is distributed within the target volume T itself (i.e., T is the same as S). For example, we want to know the radiation dose to the liver when it is distributed in the liver itself. Here, the source (liver) is also the target (liver). In such cases, the absorbed fraction is known as the self-absorbed fraction and is expressed simply as φi.
Once φi(T←S) is known, the rate of energy absorption by a target volume T is simply obtained by multiplying the rate of energy emission of radiation i (from step 1 = 213niEi) by the absorbed fraction φi(T←S), or rate of energy absorption by the target volume from radiation i = 213niEi × φi(T←S)
If there are n radiations, the rate of the total energy absorption will be equal to the sum of the energies absorbed from each radiation, that is,
The above expression can be written in a concise form as
where
is the sum of all terms when i changes from 1 to n.
How does one determine φ
i(T←S)? Determination of the absorbed fraction requires the exact knowledge of the interaction of radiation with matter, discussed in the previous chapter. Computations of φ
i(T←S) for a number of source and target combinations, from these basic mechanisms of interaction of radiation with matter, are quite involved and require the use of large computers. The
Journal of Nuclear Medicine published a variety of tables listing φ
i(T←S) for different x- or γ-ray energies and source and target volumes. As an illustration,
Table 7.2 lists the absorbed fraction for various organs of a standard man for different x- or γ-ray energies when the radionuclidic distribution is within the same organ (i.e., T is the same as S). For other combinations, the reader is referred to the original articles.
*
General Comments on φi(T←S). The maximum value of φi(T←S) can only be 1. This occurs when all emitted energy is absorbed in the target. The minimum value is 0 and occurs when there is no absorption of energy in the target.
In the case of particulate radiations such as β particles, conversion electrons, or α particles, almost all energy emitted by a radionuclide is absorbed in the volume of distribution itself, provided the source volume is larger than 1 cm3. Then, φi(T←S) = 0, unless T and S are the same, in which case φi = 1. The same holds true for x- or γ-radiation with energies less than 10 keV. Thus, these radiations deliver radiation doses only in the volume of distribution and not outside of it.
For x- or γ-radiations with energies higher than 10 keV, the absorbed fraction φ
i(T←S) strongly depends on the energy of the x- or γ-ray; the shape and size of the source volume; and the shape, size, and distance of the target volume. In general, φ
i first decreases with an increase in the energy of the x- or γ-ray and then eventually levels off. Notice in
Table 7.2 that the absorbed fraction, φ
i drops sharply with energy up to 100 keV but then does not change significantly from 100 to 500 keV for various organs listed.
Step 3—Dose Rate, dD/dt. If one now divides the rate of energy absorption by the target with its mass M, this will give the rate of energy absorption per gram of tissue, which when divided by
100 (to convert erg/g to rad) yields the dose rate for each microcurie of activity; that is, the dose rate,
By defining Δi = 2.13niEi, this reduces to
If the source volume contains A(t) µCi at time t, then the dose rate dD from A(t) amount of radioactivity becomes
Step 4—Average Dose, D. The radioactivity A(t) localized in an organ is generally a fraction f of the administered dosage A0 and is being continuously eliminated, with an effective half-life of T1/2 (eff); that is,
Therefore, the dose rate dD/dt is continuously decreasing with time and eventually becomes zero. How does one compute the total dose to the patient from the time of administration (t = 0) to the time when the dose rate has finally been reduced to zero? For this, one has to integrate the dose rate dD/dt from 0 to ∝ time, or
(dD/dt)dt. This involves integration of A(t) that, for a simple case such as
equation (7.3), leads to the following expression:
In the case where the target and source volume are the same, the self-dose D is given by
Note that the factor (fA0/M) in the above equations is the concentration of the radioactivity in the organ of localization. Therefore, it is the concentration and not the total amount of the radioactivity in an organ that is the primary determinant of the radiation dose.
From these expressions, it is evident that to minimize the radiation dose to a patient, it is desired to have smaller amounts of radioactivity (A0), radiopharmaceuticals with shorter T1/2 (eff), and radionuclides with smaller absorbed fractions (which means γ-ray energy higher than 100 keV and no particulate emission as discussed earlier; p. 43).