Internal Radiation Dosimetry

Absorption of energy from ionizing radiation can cause damage to living tissues. This is used to advantage in radionuclide therapy, but it is a limitation for diagnostic applications because it is a potential hazard for the patient. In either case, it is necessary to analyze the energy distribution in body tissues quantitatively to ensure an accurate therapeutic prescription or to assess potential risks. The study of radiation effects on living organisms is the subject of radiation biology (or radiobiology ) and is discussed in several excellent texts, some of which are listed at the end of this chapter.

One of the most important factors to be evaluated in the assessment of radiation effects on an organ is the amount of radiation energy deposited in that organ. Calculation of radiation energy deposited by internal radionuclides is the subject of internal radiation dosimetry. There are two general methods by which these calculations may be performed: the classic method and the absorbed fraction method. Although the classic method is somewhat simpler, and the results by the two methods are not greatly different, the absorbed fraction method (also known generally as the MIRD method , after the Medical Internal Radiation Dose Committee of the Society of Nuclear Medicine) is more versatile and gives more accurate results. Therefore it has gained wide acceptance as the standard method for performing internal dosimetry calculations. The procedures to be followed in using the absorbed fraction method are summarized in this chapter. Dosimetry calculations for external radiation sources as well as health physics aspects of radiation dosimetry are discussed in Chapter 23 . Some radiation dose estimates for nuclear medicine procedures are summarized in Appendix E .

A

Radiation Dose and Equivalent Dose: Quantities and Units

Radiation dose, D, refers to the quantity of radiation energy deposited in an absorber per gram of absorber material. This quantity applies to any kind of absorber material, including body tissues. The basic unit of radiation dose is the gray , abbreviated Gy * :

* This unit is named after Harold Gray, a British medical physicist best known for his discovery of the “oxygen effect” in radiation therapy.

1 Gy = 1 joule energy deposited per kg absorber

$1 Gy = 1 joule energy deposited per kg absorber$

The traditional unit for absorbed dose is the rad, an acronym for radiation absorbed dose:

1 rad = 100 ergs energy deposited per g absorber

$1 rad = 100 ergs energy deposited per g absorber$

Since 1 joule = 10 ⁷ ergs, 1 Gy is equivalent to 100 rads or, alternatively, 1 rad = 10 ^–2 Gy = 1 cGy. As is the case for units of activity, progress in the transition from traditional to SI units varies with geographic location, with SI units dominating practice in Europe, whereas traditional units still are commonplace in the United States. In this chapter, radiation doses are presented in grays, with values in rads also indicated in selected examples.

Equivalent dose, symbolically indicated by H _T , is a quantity that takes into account the relative biologic damage caused by radiation interacting with a particular tissue or organ. Tissue damage per gray of absorbed dose depends on the type and energy of the radiation, and how exactly the radiation deposits its energy in the tissue. For example, an α particle has a short range in tissue and deposits all of its energy in a very localized region. In contrast, γ rays and electrons deposit their energy over a wider area. Table 22-1 shows the radiation weighting factors, w _R , used to calculate equivalent dose for different types and energies of radiation. The SI unit of equivalent dose is the sievert * (Sv). It is related to the average absorbed dose D in an organ or tissue, T, by

TABLE 22-1

WEIGHTING FACTORS FOR DIFFERENT TYPES OF RADIATION IN THE CALCULATION OF EQUIVALENT DOSE ^†

Type of Radiation	Radiation Weighting Factor, w _R
x rays	1
γ rays	1
Electrons, positrons	1
Neutrons	Continuous function of neutron energy
Protons >2 MeV	2
α particles, fission fragments, heavy ions	20

† Data from reference .

* This unit is named after Rolf M. Sievert, best known for his development of elaborate mathematical models, including the Sievert integral, which for many years provided the basis for calculating radiation doses from implanted radium needles. He also constructed and performed many basic measurements with ionization chambers.

H_{T} = D_{T} \times w_{R}

$H_{T} = D_{T} \times w_{R}$

Equivalent dose replaces an older quantity known as the dose equivalent. The dose equivalent is based on the absorbed dose at a point in an organ (rather than an average across the whole organ) and is weighted by quality factors, Q , that are similar to w _R . The unit for dose equivalent also is the Sv.

The traditional unit for dose equivalent is the roentgen-equivalent man (rem). The conversion factor between traditional and SI units is

1 rem = 10^{- 2} Sv = 1 cSv = 10 mSv

$1 rem = 10^{- 2} Sv = 1 cSv = 10 mSv$

For radiations of interest in nuclear medicine (γ rays, x rays, electrons, and positrons) the radiation weighting factor is equal to 1. Therefore the equivalent dose or dose equivalent in Sv (or rem) is numerically equal to the absorbed dose in Gy (or rads).

B

Calculation of Radiation Dose (MIRD Method)

1 Basic Procedure and Some Practical Problems

The absorbed fraction dosimetry method allows one to calculate the radiation dose delivered to a target organ from radioactivity contained in one or more source organs in the body ( Fig. 22-1 ). The source and target may be the same organ, and, in fact, frequently the most important contributor to radiation dose is radioactivity contained within the target organ itself. Generally, organs other than the target organ are considered to be source organs if they contain concentrations of radioactivity that exceed the average concentration in the body.

$FIGURE 22-1, Absorbed dose delivered to a target organ from one or more source organs containing radioactivity is calculated by the absorbed fraction dosimetry method.$

The general procedure for calculating the radiation dose to a target organ from radioactivity in a source organ is a three-step process, as follows:

1
The amount of activity and time spent by the radioactivity in the source organ are determined. Obviously, the greater the activity and the longer the time that it is present, the greater is the radiation dose delivered by it.
2
The total amount of radiation energy emitted by the radioactivity in the source organ is calculated. This depends primarily on the energy of the radionuclide emissions and their frequency of emission (number per disintegration).
3
The fraction of energy emitted by the source organ that is absorbed by the target organ is determined. This depends on the type and energy of the emissions (absorption characteristics in body tissues) and on the anatomic relationships between source and target organs (size, shape, and distance between them).

Each of these steps involves certain difficulties. Step 2 involves physical characteristics of the radionuclide, which generally are known accurately. Step 3 involves patient anatomy, which can be quite different from one patient to the next. Step 1 is perhaps the most troublesome. Such data on radiopharmaceutical distribution as are available usually are obtained from studies on a relatively small number of human subjects or animals. There are variations in metabolism and distribution of radionuclides among human subjects, especially in different disease states. Also, the distribution of radioactivity within an organ may be inhomogeneous, leading to further uncertainties in the dose specification for that organ.

Because of these complications and variables, radiation dose calculations are made for anatomic models that incorporate “average” anatomic sizes and shapes. The radiation doses that are calculated are average values of D for the organs in this anatomic model. An exception is made when one is specifically interested in a surface dose to an organ from activity contained within that organ, for example, the dose to the bladder wall resulting from bladder contents. This is considered to have a value one-half the average dose to the organ or, in this case, the bladder contents.

In spite of the refined mathematical models used in the absorbed fraction model, the results obtained are only estimates of average values. Thus they should be used for guideline purposes only in evaluating the potential radiation effects on a patient.

2 Cumulated Activity,

The radiation dose delivered to a target organ depends on the amount of activity present in the source organ and on the length of time for which the activity is present. The product of these two factors is the cumulated activity in the source organ. The SI unit for cumulated activity is the becquerel • sec (Bq • sec). The corresponding traditional unit is the µCi • hr (1 µCi = 3.7 × 10 ⁴ Bq; 1 hr = 3600 sec; therefore, 1 µCi • hr = 3.7 × 10 ⁴ × 3600 = 1.332 × 10 ⁸ Bq • sec = 1.332 × 10 ² MBq • sec). Cumulated activity is essentially a measure of the total number of radioactive disintegrations occurring during the time that radioactivity is present in the source organ. The radiation dose delivered by activity in a source organ is proportional to its cumulated activity.

Each radiotracer has its own unique spatial and temporal distribution in the body, as determined by radiotracer delivery, uptake, metabolism, clearance and excretion, and the physical decay of the radionuclide. The amount of activity contained in a source organ therefore generally changes with time. If the time-activity curve is known, the cumulated activity for a source organ is obtained by measuring the area under this curve ( Fig. 22-2 ). Mathematically, if the time-activity curve is described by a function A ( t ), then

\tilde{A} \approx \int_{0}^{\infty} A (t) d t

$\tilde{A} \approx \int_{0}^{\infty} A (t) d t$

FIGURE 22-2, Hypothetical time-activity curve for radioactivity in a source organ. Cumulated activity in Bq • sec is the area under the curve (equivalent to the integral in Equation 22-5 ).

where it is assumed that activity is administered to the patient at time t = 0 and is measured to complete disappearance from the organ ( t = ∞).

To estimate the radiation dose received from a particular radiotracer, time-activity curves for all the major organs are required. These can be obtained from animal studies (which are then extrapolated with some uncertainty to the human), imaging studies in normal human subjects, prior knowledge of the tracer kinetics, or some combination of these approaches. Time-activity curves can be quite complex, and thus Equation 22-5 may be difficult to analyze. Frequently, however, certain assumptions can be made to simplify this calculation.

Situation 1: Uptake by the organ is “instantaneous” ( i.e., very rapid with respect to the half-life of the radionuclide ) , and there is no biologic excretion. The time-activity curve then is described by ordinary radioactive decay ( Equation 4-7, Equation 4-10 ):

A (t) = A_{0} e^{- 0.693 t / T_{p}}

$A (t) = A_{0} e^{- 0.693 t / T_{p}}$

where T _p is the physical half-life of the radionuclide and A ₀ is the activity initially present in the organ. Thus

\begin{matrix} \tilde{A} \approx A_{0} \int_{0}^{\infty} e^{- 0.693 t / T_{p}} d t \\ = \frac{T_{p} A_{0}}{0.693} = 1.44 T_{p} A_{0} \end{matrix}

$\begin{matrix} \tilde{A} \approx A_{0} \int_{0}^{\infty} e^{- 0.693 t / T_{p}} d t \\ = \frac{T_{p} A_{0}}{0.693} = 1.44 T_{p} A_{0} \end{matrix}$

The quantity 1.44 T _p is the average lifetime of the radionuclide (see Chapter 4 , Section B.3). Thus the cumulated activity in a source organ, when eliminated by physical decay only, is the same as if activity were present at a constant level A ₀ for a time equal to the average lifetime of the radionuclide ( Fig. 22-3 ).

FIGURE 22-3, Illustration of relationship between and average lifetime (1.44 T p ) of a radionuclide for simple exponential decay.

Example 22-1

What is the cumulated activity in the liver for an injection of 100 MBq of a ^99m Tc-labeled sulfur colloid, assuming that 60% of the injected colloid is trapped by the liver and retained there indefinitely?

Answer

\begin{matrix} \tilde{A} = 1.44 \times 100 MBq \times 0.60 \times 6.0 hr \\ = 518.4 MBq \cdot hr \\ = 1.87 \times 10^{6} MBq \cdot sec \end{matrix}

$\begin{matrix} \tilde{A} = 1.44 \times 100 MBq \times 0.60 \times 6.0 hr \\ = 518.4 MBq \cdot hr \\ = 1.87 \times 10^{6} MBq \cdot sec \end{matrix}$

Situation 2: Uptake is instantaneous, and clearance is by biologic excretion only ( no physical decay, or physical half-life very long in comparison with biologic excretion ). In this situation, biologic excretion must be carefully analyzed. Frequently, it can be described by a set of exponential excretion components, with a fraction f ₁ of the initial activity A ₀ being excreted with a (biologic) half-life T _b1 , a fraction f ₂ with half-life T _b2 , and so on ( Fig. 22-4 ). The cumulated activity then is given by

\begin{matrix} \tilde{A} \approx A_{0} \int_{0}^{\infty} f_{1} e^{- 0.693 t / T_{b 1}} d t \\ + A_{0} \int_{0}^{\infty} f_{2} e^{- 0.693 t / T_{b 2}} d t + \dots \\ = 1.44 T_{b 1} f_{1} A_{0} + 1.44 T_{b 2} f_{2} A_{0} + \dots \end{matrix}

$\begin{matrix} \tilde{A} \approx A_{0} \int_{0}^{\infty} f_{1} e^{- 0.693 t / T_{b 1}} d t \\ + A_{0} \int_{0}^{\infty} f_{2} e^{- 0.693 t / T_{b 2}} d t + \dots \\ = 1.44 T_{b 1} f_{1} A_{0} + 1.44 T_{b 2} f_{2} A_{0} + \dots \end{matrix}$

FIGURE 22-4, Illustration of a multicomponent exponential excretion curve. Fraction f 1 is excreted with biologic half-life T b1 , f 2 with half-life T b2 , f 3 with half-life T b3 , and so on.

Example 22-2

Suppose that 100 MBq of ^99m Tc-labeled microspheres are injected into a patient, with essentially instantaneous uptake of activity by the lungs. What is the cumulated activity in the lungs if 60% of the activity is excreted from the lungs with a biologic half-life of 15 minutes and 40% with a biologic half-life of 30 minutes?

Answer

Because ^99m Tc physical decay is much slower than the biologic excretion process, we may assume that no physical decay occurs during the time that activity is present in the lungs. Thus ( Equation 22-8 )

\begin{matrix} \tilde{A} = (1.44 \times 1 / 4 hr \times 0.60 \times 100 MBq) \\ + (1.44 \times 1 / 2 hr \times 0.40 \times 100 MBq) \\ = (21.6 + 28.8) MBq \cdot hr \\ = 50.4 MBq \cdot hr \\ = 1.81 \times 10^{5} MBq \cdot sec \end{matrix}

$\begin{matrix} \tilde{A} = (1.44 \times 1 / 4 hr \times 0.60 \times 100 MBq) \\ + (1.44 \times 1 / 2 hr \times 0.40 \times 100 MBq) \\ = (21.6 + 28.8) MBq \cdot hr \\ = 50.4 MBq \cdot hr \\ = 1.81 \times 10^{5} MBq \cdot sec \end{matrix}$

Situation 3: Uptake is instantaneous but clearance by both physical decay and biologic excretion are significant . In this case, if biologic excretion is described by a single-component exponential curve with biologic half-life T _b , and the physical half-life is T _p , then the total clearance is described by a single-component exponential curve with an effective half-life T _e given by *

* Equation 22-9 can be derived from Equation 4-2 , Equation 4-9 by treating biologic excretion as the equivalent of a second pathway in a “branching” radioactive decay scheme.

\frac{1}{T_{e}} = \frac{1}{T_{p}} + \frac{1}{T_{b}}

$\frac{1}{T_{e}} = \frac{1}{T_{p}} + \frac{1}{T_{b}}$

T_{e} = \frac{T_{p} T_{b}}{(T_{p} + T_{b})}

$T_{e} = \frac{T_{p} T_{b}}{(T_{p} + T_{b})}$

Cumulated activity is given by

\tilde{A} \approx 1.44 T_{e} A_{0}

$\tilde{A} \approx 1.44 T_{e} A_{0}$

If there is more than one component to the biologic excretion curve, then each component has an effective half-life given by Equation 22-9 for that component, and the cumulated activity is computed with effective half-lives replacing biologic half-lives in Equation 22-8 .

Example 22-3

Suppose in Example 22-2 that because of a metabolic defect 60% of the activity is excreted from the lungs with a half-life of 2 hours and 40% with a half-life of 3 hours. What is the cumulated activity in the lungs for a 100 MBq injection for this patient?

Answer

The effective half-lives for the two components of biologic excretion are ( Equation 22-10 )

T_{e 1} = 6 \times 2 / (6 + 2) = 1.5 hr

$T_{e 1} = 6 \times 2 / (6 + 2) = 1.5 hr$

T_{e 2} = 6 \times 3 / (6 + 3) = 2 hr

$T_{e 2} = 6 \times 3 / (6 + 3) = 2 hr$

Thus applying Equation 22-8 , with T _e replacing T _b ,

\begin{matrix} \tilde{A} = (1.44 \times 1.5 hr \times 0.60 \times 100 MBq) \\ + (1.44 \times 2 hr \times 0.40 \times 100 MBq) \\ = (129.6 + 115.2) MBq \cdot hr \\ = 244.8 MBq \cdot hr \\ = 8.81 \times 10^{5} MBq \cdot sec \end{matrix}

$\begin{matrix} \tilde{A} = (1.44 \times 1.5 hr \times 0.60 \times 100 MBq) \\ + (1.44 \times 2 hr \times 0.40 \times 100 MBq) \\ = (129.6 + 115.2) MBq \cdot hr \\ = 244.8 MBq \cdot hr \\ = 8.81 \times 10^{5} MBq \cdot sec \end{matrix}$

Situation 4: Uptake is not instantaneous. The equations developed thus far will overestimate radiation doses when uptake by the source organ is not rapid in comparison with physical decay, that is, if a significant amount of physical decay occurs during the uptake process, before the activity reaches the source organ of interest. This situation arises with radionuclides that have a slow pattern of uptake in comparison with their physical half-life. Frequently, uptake can be described by an exponential equation of the form

A (t) = A_{0} (1 - e^{- 0.693 t / T_{u}})

$A (t) = A_{0} (1 - e^{- 0.693 t / T_{u}})$

where T _u is the biologic uptake half-time. In this case, cumulated activity is given by

\tilde{A} \approx 1.44 A_{0} T_{e} (T_{ue} / T_{u})

$\tilde{A} \approx 1.44 A_{0} T_{e} (T_{ue} / T_{u})$

where T _e is the effective excretion half-life ( Equation 22-10 ) and T _ue is the effective uptake half-time

T_{ue} = \frac{T_{u} T_{p}}{T_{u} + T_{p}}

$T_{ue} = \frac{T_{u} T_{p}}{T_{u} + T_{p}}$

Example 22-4

A radioactive gas having a half-life of 20 seconds is injected in an intravenous solution. It appears in the lungs with an uptake half-time of 30 seconds and is excreted (by exhalation) with a biologic half-life of 10 seconds. What is the cumulated activity in the lungs for a 250-MBq injection?

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A

Radiation Dose and Equivalent Dose: Quantities and Units

B

Calculation of Radiation Dose (MIRD Method)

1

Basic Procedure and Some Practical Problems

2

Cumulated Activity,

Example 22-1

Answer

Example 22-2

Answer

Example 22-3

Answer

Example 22-4

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