This chapter is arranged in three parts:
-
List of the Information Needed to Calculate the Number of
Tracks per Nucleus, p.1
Provision of the Input, the Calculations, and the Answer,
p.2
The Fallacy of Slow Delivery of Very Low Doses, p.6
Then tables.
Readers of Chapter 18 already know how the disproof of any safe dose or dose-rate is related to the approximate number of primary ionization tracks occurring in cell-nuclei.
In this chapter, we will show step-by-step how we determined the number of tracks per nucleus which are occurring at any particular dose (for instance, one rad or centi-gray), from a particular radiation source. Separately, in Chapter 33, we will show that our method and the methods of other analysts must be in good agreement, because they produce closely similar results.
1. List of the Requisite Information
In order to find out how many primary electron tracks traverse a nucleus at a tissue-dose of one rad, (and how many rads correspond with an average of one track per nucleus), we will need the pieces of information described below. Although the average number of tracks per nucleus, at one rad, comes from Item I (Eye) divided by Item J, we cannot skip the earlier items because each of them is required in order to obtain the value for Item Eye.
-- (A) Definition of a rad
(or centi-gray). A dose of
1 centi-gray (1 rad) of low-LET radiation is, by
definition, the deposition of 10^-5 joules, or 6.24 x
10^10 KeV, per gram of tissue.
-- (B) Average KeV per photon. We will consider
photons from medical X-rays, radium-226 (and
daughters), cesium-137, and A-bomb gamma rays.
-- (C) Number of photons whose energy must be
totally absorbed in one gram of tissue to produce a dose
of one centi-gray (cGy).
-- (D) Number and energy of the high-speed
electrons produced per photon. The term
"electron-packet" will be defined.
-- (E) Distance across a typical human cell.
-- (F) Distance traveled by each high-speed
electron. Relationship to range (see Chapter 33, p.10).
-- (G) Number of cells traversed by each electron
in an electron- packet.
-- (H) The total number
of cell-traversals made
by all the primary electrons required to deliver a
tissue-dose of 1 centi-gray to 1 gram. Photo-electrons
and Compton electrons are both "primary" electrons, in
contrast to secondary electrons produced along a
primary electron track.
-- (Eye) The total number of
nuclear-traversals
made by all the primary electrons required to deliver a
dose of 1 cGy to 1 gram of tissue.
-- (J) The total number of cell-nuclei which are
available to be traversed in a gram of tissue.
-- (K) The average number of primary ionization
tracks through each nucleus at a tissue-dose of 1 cGy
(1 rad). This item is (Eye / J).
-- (L) The tissue-dose when the average
track-rate per nucleus = 1.0.
-- (M) An "If . . . Then" table showing
correspondences between tracks and various doses.
When we have the rate of tracks per nucleus at 1 cGy,
we can readily tabulate the average number of tracks
per nucleus at any dose above or below 1 cGy, since the
number of tracks is proportional to dose^1 for a specific
radiation source. In addition, we can start with an
average of one track (or any other number of tracks) per
nucleus, and calculate the tissue-dose associated with
that number of tracks.
-- (N) The
Poisson distribution of tracks. When the
average number of tracks per nucleus is 1.0, some
nuclei will have no tracks at all, and some will have
more than one track. We shall examine the distribution
of tracks by using the Poisson equation. Readers have
already seen (Chapter 18,
Part 6)
that, after the average number of tracks per nucleus has been
determined for a specified dose, the Poisson equation
tells the probability of various numbers of tracks per
nucleus.
2. Provision of the Input, the Calculations, and the Answer
We shall take up the topics in the alphabetical order above. Items F and G are handled together. Because Topics A, B, E, and J require no tables, the reader will find no Tables 20-A, 20-B, 20-E, or 20-J at the end.
-- (A) Definition of a Rad or Centi-Gray (cGy) :
One rad or cGy means 6.24 x 10^10 KeV of energy per gram of tissue.
-- (B) Average KeV per Photon :
This analysis will consider photons of four different energy regions.
(1) 30 KeV X-Rays :
These X-rays are the most likely ones to characterize the medical exposures in eight of the nine epidemiological studies summarized in Chapter 21 (and in Table 21-A). When peak kilovoltage across an X-ray tube is 90, the average energy per photon is about 30 KeV.
(2) Radium-226 and Its Daughters :
These gamma rays are the source of exposure in one of the nine studies in Table 21-A (the British Luminizers). The estimated energy per average photon is 596 KeV.
(3) Cesium-137 Gamma Rays :
Cesium-137 is the principal source of population-exposure from the Chernobyl nuclear power accident; the gamma ray is actually from barium-137m decaying to barium-137, and the estimated energy is 662 KeV per photon. Although Chernobyl is not part of our disproof of any safe dose or dose-rate, some readers may wish to know how cesium-137 compares with radium-226 and medical X-rays, so it is evaluated here for reference.
(4) Atomic-Bomb Gamma Rays at Hiroshima :
Radiation from the Hiroshima and Nagasaki atomic bombs included quite a mixture of energies. Since the study of A-bomb survivors is not one of our nine studies in Chapter 21, we need only a "ballpark" estimate of the KeV per average photon. I have done a very simplified estimate based on Hiroshima data in the DS86 system. The value we shall use is 1608 KeV per average photon. (See Chapter 32 for calculation of this value.)
From Photons to Primary Electrons :
In order to determine values for Item D below, we will have to take account of the three ways in which high-energy photons deliver energy to the molecules of cellular tissue.
Photo-Electric Effect. The photo-electric effect means that the photon disappears and a single electron is set into high-speed motion. The electron carries off all the energy of the photon minus a much smaller amount of energy required to lift the electron out of the atom (Par87, p.93). We will disregard the latter. The photo-electric effect is dominant for photon energies below about 40 KeV, in materials of low atomic number such as cellular tissue (Par87, p.93).
Compton Effect. In the Compton effect, which dominates at energies above those where the photo-electric effect dominates, only part of the energy of the photon is transferred to set an electron into high-speed motion, and the remainder of the energy is carried off by a new photon. This new photon, if its energy is sufficient, can then again participate in what is commonly called a Compton process: Setting an electron in motion and again creating a new photon of further reduced energy. Finally, when these reduced-energy photons have energies where the photo-electric effect dominates, the remaining energy is transferred in toto to an electron.
Pair Production. If the photon's energy is above 1.02 MeV, another type of interaction is possible, namely disappearance of the original photon, its energy being distributed as follows: 1.02 MeV is converted into two particles, an electron and a positron, and the remaining energy goes into energy of motion of the electron and the positron. Since photons cannot create the positron-electron pair at any energy below 1.02 MeV, pair production is irrelevant for the gamma rays of radium-226, cesium-137, and for the low energy X-rays, which are of major interest in this analysis. For part of our analysis, which deals with the gamma rays from A-bomb radiation where pair production is possible, we shall make the approximation of neglecting pair production in comparison with the other processes described above. Paretzke agrees that for gamma energies below 2 MeV, pair production can be neglected compared with the Compton process (Par87, p.95).
Coherent Scattering. There is a fourth process of interaction of photons with tissue, known as coherent (Rayleigh) scattering. No energy is delivered to tissue, but the direction of the photon is changed. This process becomes prominent at energies below about 0.1 MeV. Because there is no energy transferred, we will not need to consider coherent scattering in our analyses.
-- (C) Number of Photons Required to Deliver 1 Rad to 1 Gram :
In this analysis, we are not at all concerned with photons which pass right through a tissue without converting to high-speed electrons. Such photons contribute no dose to the tissue. We are concerned with the question: When a tissue receives a dose of one rad (one centi-gray), how many photons did convert to high-speed electrons in order to deliver that dose to a single gram of tissue?
The answer comes from dividing the energy-deposition required (Item A), by the average energy supplied by each photon (Item B). Table 20-C provides the answers for the four types of radiation which we are evaluating.
While we are considering large-area radiation with essentially total absorption of the energy of the initial photon (if it interacts at all), we have given attention to the implication for our results if some post-Compton photons are lost from tissue. See Chapter 33 for these considerations.
-- (D) Electrons per Photon, and "Packets" Defined :
For the medical X-rays, the photo-electric effect is overwhelmingly dominant, so we are fully justified in stating that there will be one high-speed electron produced per 30-KeV photon, and that it carries all of the photon's energy (except for the very small binding energy of the electron).
For the radium, cesium, and A-bomb gamma rays, no such simplification would be realistic because each photon produces several high-speed electrons of successively lower energy by Compton processes. We shall call the whole set, produced from a single photon, its "packet" of high-speed or primary electrons. Since each electron in a packet produces its own primary ionization track, we must take account of each electron.
The calculations are presented in Chapter 32; the results have been transferred forward into Table 20-D.
-- (E) Distance across a Typical Human Cell, Cuboid Model :
The choice of appropriate cell-size is based on measurements from 38 electron micrographs of normal human cells (Elias78; Gar76; Ham85; Jo85). The mean nuclear diameter was 5.9 micrometers for 29 non-fetal human cells; 6.1 micrometers for 6 fetal cells; 5.7 micrometers for one non-fetal thyroid cell; 6.9 micrometers for one fetal thyroid cell; and 5.5 micrometers for one non-fetal breast cell.
To the weighted average of 5.9 micrometers, two corrections were made. Because it was impossible to know that the nuclei pictured were cut exactly through the maximum dimension, a factor of 1.1 increase was applied. Because it was possible that fixation of the tissue may have caused some shrinkage of cells, another factor of 1.1 increase was applied. With these corrections to the observations, the diameter of nuclei taken for this analysis is 7.1 micrometers -- as it was in Go86.
A very reasonable estimate, from examination of numerous histology texts, is that cell diameter is twice the nuclear diameter, or 14.2 micrometers.
With regard to our nuclear diameter of 7.1 micrometers, it is interesting to note that others are now using similar values for similar purposes. For instance, Brackenbush and Braby use a nuclear diameter of 7.0 micrometers. In a recent discussion of the microdosimetric basis for exposure limits, they state:
"Since most biological effects appear to be the consequence of the autonomous response of individual cells, the frequency and magnitude of the events in cells is pertinent. If we consider a 7 micrometer-diameter sphere as typical of a cell nucleus, we can estimate this frequency" (Brack88, p.252). Readers may note that Brack88 goes directly to the nucleus as the relevant site in the cell.
Goodhead (Good88, p.237) uses a value of 7.5 micrometers for the typical nuclear diameter. Neither Brack88 nor Good88 provides the basis for the value chosen.
Adjusted Values for a Cuboidal Model :
For simplification of this analysis, we are going to treat the cells and their nuclei as though they are cuboidal rather than spherical. This treatment will not result in any major changes in our expectations of cells traversed by tracks, and the mathematics are grossly simplified.
A spherical nucleus with a diameter of 7.1 micrometers has the same volume as a cuboidal nucleus of 5.7 micrometers per edge of the cube. A spherical cell with a diameter of 14.2 micrometers has the same volume as a cuboidal cell of 11.4 micrometers per edge.
For further simplification, we are going to treat the radiation source as one which is normal (perpendicular) to one face of the cuboidal cells. With the data for cuboidal cells and the approximation that all the photo-electrons and Compton electrons come in perpendicular to one face of all the cuboidal cells, we can proceed to the analysis of the tracks per cell and tracks per nucleus from the various radiations of interest.
-- (F) Distance Traveled by
Each High-Speed Electron (Range),
and
-- (G) Number of Cells Traversed by
Each Electron in a Packet :
The derivation of the ranges for all primary electrons is shown in detail in Chapter 33. The answers from Chapter 33 have been brought forward here into Table 20-FG, as Item F.
Item G, the number of cells traversed by each electron, is obviously the electron's range in micrometers, divided by 11.4 micrometers (the depth of the cuboidal cell).
-- (H) Total Cell-Traversals
by Tracks Delivering 1 Rad to 1 Gram :
The next step is the determination of the total number of cell-traversals (by primary electrons) which occur when 1 rad is delivered to 1 gram of tissue. In Table 20-H, the total number of cell-traversals per rad of tissue-dose is presented for each type of radiation. This is, of course, the number of photons required to deliver one rad to one gram (Item C), times the total cells traversed by all the primary electrons produced by such photons (Item G, sum).
Total number of cell-traversals does not mean number of different cells traversed. Some cells experience multiple traversals.
As stated in Item C (text), our treatment is based on the approximation, which we consider reasonable for large-area irradiation, that for each original photon which undergoes the Compton process, all the energy of the post-Compton photon is also converted to electron energy by successive processes in the tissue.
-- (Eye) Total Nuclear-Traversals
by Tracks Delivering 1 Rad to 1 Gram :
In our model of the cuboidal cells and cuboidal nuclei, the area of one face of the nucleus is 1/4 of the area of the one face of the whole cell, since the edge length of the nucleus is 1/2 of that of the whole cell. Thus, for electrons normal to the cells, the nuclear area "seen" by the electrons is 1/4 of the cellular area "seen." Therefore the nuclear traversals are going to be the cellular traversals (Item H) times (0.25). These values are presented in Table 20-Eye.
Total number of nuclear-traversals does not mean number of different nuclei traversed. Some nuclei experience multiple traversals.
-- (J) Number of Nuclei Available
for Traversal in 1 Gram of Tissue :
We need to know how many nuclei are present and available for traversal, in one gram of tissue. Number of nuclei = number of cells. We will know the number of cells present in one gram of tissue if we divide (volume of one gram of cells) / (volume of a single cell).
Volume of One Gram of Cells: At an approximate density of 1.0 gram per cm^3, the volume of one gram of cells is 1.0 cm^3 (one cubic centimeter). And one cm^3 represents 10^12 micrometers^3 (one trillion cubic microns).
Volume of a Single Cell: For a cuboidal cell, 11.4 micrometers on an edge, the volume is (11.4 micrometers)^3, or 1481.544 micrometers^3.
So, number of nuclei per gram of cells = (10^12 / 1481.544) = 6.75E+08 nuclei per gram of cells, or about 675 million.
-- (K) Average Number of Primary Tracks
Traversing a Nucleus at 1 Rad :
We need only divide the total number of nuclear-traversals which are occurring (Item Eye), by the total number of nuclei which are available for traversal in a gram of tissue (Item J), in order to determine how many tracks are traversing each nucleus, on the average, at a tissue-dose of 1 rad (cGy). The results are provided in Table 20-K.
Comparison with Other Estimates: Is there a disparity between our estimates in Table 20-K, and estimates recently made by some other analysts? The short answer is, "no." What may seem like differences are reconciled in Chapter 33. It looks, perhaps, as if not everyone is taking the Compton process into account yet.
Variation in Tissues: When we did this type of analysis earlier (Go86), we had to ask ourselves a question which some readers may be asking themselves, too: "Are the values in Table 20-K valid even where the number of nuclei per gram might vary, due to the presence of connective tissue, nerves, interstitial fluid, and such things?" The short answer is, "yes."
The ratio (tracks per nucleus at 1 cGy or rad) would not be altered if there were fewer nuclei per gram of tissue, due to the presence of connective tissue, interstitial fluid, and so forth. Likewise, the ratio is not altered when cells which do not produce cancer -- such as nerve and muscle cells -- are part of the irradiated gram of tissue. The volume so occupied can be regarded as if it were all occupied by cells containing no relevant nuclei. For instance, if there are 25 % fewer nuclei in a gram, then Item J would become (0.75) x (nuclei present). Likewise, Item Eye would become (0.75) x (nuclear traversals). When Item Eye is divided by Item J, the effect cancels out, and the ratio of average tracks per nucleus remains the same.
-- (L) Tissue-Dose When the
Average Track-Rate per Nucleus Is One :
Because we know, from Table 20-K, the rate of nuclear tracks per rad, it follows that we also know the rate of rads per nuclear track. As a convenience, Table 20-L provides the computed values for each of the four types of radiation which we have examined.
-- (M) "If . . . Then" Table,
Showing Corresondence between Tracks & Doses :
As a convenient reference, Table 20-L uses the basic ratios (of average tracks per nucleus) from Table 20-K to compute, "If the total tissue-dose is X, then the average number of tracks per nucleus is Y." And in reverse, "If the average number of tracks per nucleus is Y, then the tissue-dose is X."
Readers who compare the entries for 30 KeV X-rays, with the entries for A-bomb gamma rays, will notice that the A-bomb electron-tracks have to traverse about four-fold more nuclei in order to deliver the same amount of dose (for instance, 1 rad). In other words, electron-tracks from the medical X-rays deliver the same amount of energy in a shorter linear range. They pack the energy-transfers more densely, on the whole.
This observation is related to our warning, that the cancer-hazard from medical X-rays may be underestimated by the A-Bomb Study (Chapter 13, Part 4). It is widely thought that the biological menace of ionizing radiation (its RBE, or Relative Biological Effectiveness) rises with the density of its energy-depositions. Indeed, in Chapter 13, we cite estimates that the RBE of 250 kVp X-rays may be two, compared with high-energy gamma rays.
-- (N) Poisson Distribution of Tracks
In an irradiated tissue, either a nucleus is "hit" by one or more tracks, or it is not hit at all.
When events occur independently of each other, as tracks do, we can use Poisson statistics to determine the chance of getting zero, 1, 2, 3, 4, 5, etc., events (tracks) per nucleus when we know that the average is one track per nucleus -- or any other average. The equation which describes the distribution of probabilities is:
-
p(V) = {EXP(-N)} x {N^V / V!}
-- V is the number of tracks for which we
want the probability calculated. (Compare Column A
versus Column B in the tabulation.)
-- p(V) is the probability of exactly "V"
events occurring.
-- N is the average number of events: Tracks
per nucleus.
-- V! is "V factorial," which means the
product of all numbers starting with 1 and going up
through "V" -- for all V greater than zero. For
instance, for V = 5, V! = 1x2x3x4x5.
-- An important reminder: 0! = 1, NOT zero.
-- {EXP(-N)} means raising the value e (base of
natural logarithms) to the power, (-N). The value of e is
2.718281828.
Exp(N) is a widely used device (perhaps not in all countries, however) which avoids superscripts when stating that the value "e" is raised to the power "N." Of course, Exp(-N) is the same as 1/Exp(N).
The tabulation illustrates the use of the Poisson equation to estimate the probability of getting zero events and getting 5 events when the average number of events is 1 track per nucleus. For simplicity of calculations, it is useful to set up a tabular format. The desired information, p(V), is in Column G. Readers will see that Column G = (Column C x Column F) -- in harmony with the original equation with which we started.
Col. Col. Col. Col. Col. Col. Col.
A B C D E F G
-----------------------------------------------------------
V = p(V) =
Number of Probability
N = Tracks for of
Avg. Which the Exactly
Number Probability V
of Is Desired Tracks
Tracks
N V EXP(-N) N^V V! N^V/V! p(V)
-----------------------------------------------------------
1.00 0 0.367879 1 1 1 0.367879
1.00 5 0.367879 1 120 0.008333 0.003065
The entries in Column D are the value of N raised to the power, V. Thus, 1^0 = 1, and 1^5 = 1 also.
The entries in Column E are V!. Thus, 0! = 1, and 5! = 1x2x3x4x5 = 120.
We can use the equation to construct tables, like the three in Table 20-N. Each line, of each table, requires one use of the equation. Of course, the equation can be set up on a computer, with the entire calculation done in one step.
3. The Fallacy of Slow Delivery of Very Low Doses
The previous chapter made it clear why the Least Possible Disturbance to a single nucleus is traversal by one primary electron-track. However, there is no tissue-dose at which disturbance occurs uniformly in cell-nuclei. The top section of Table 20-N has used the Poisson equation to find out what the actual distribution of tracks per nucleus is, when the average frequency from a tissue-dose is one track per nucleus.
It turns out (Table 20-N) that, at a tissue-dose where the average track-frequency is one per nucleus, only 26.4 % of the nuclei in the exposed tissue "feel" more than the Least Possible Disturbance (one track). The distribution is:
36.8 % of the nuclei receive no track at all.
36.8 % of the nuclei receive exactly one track.
26.4 % of the nuclei receive two or more tracks.
-------
100.0 %
In other words, for many purposes, it would be reasonable to regard an acute tissue-dose which delivers one primary electron per nucleus, on the average, as the lowest conceivable dose and dose-rate at the level of the nucleus. What is that dose? Both Tables 20-L and 20-M provide the information, for the four types of radiation examined in this chapter.
Of course, the average number of tracks is directly proportional to dose (Items C and D). As dose falls below the level where 1.0 is the average number of tracks per nucleus, the track-average falls accordingly. With each change in the average, the distribution of tracks has to be newly calculated with the Poisson equation. The mid-section of Table 20-N shows the distribution for tissue-doses where the average is 0.05 track per nucleus:
95.1 % of the nuclei receive no track at all.
4.8 % of the nuclei receive exactly one track.
0.1 % of the nuclei receive two or more tracks.
-------
100.0 %
Correct Perception of Dose-Rate :
The two lists of percentages above can help improve the perception of dose-rate. We will use 30 KeV X-rays to illustrate what is, and is not, meant by "a lower dose-rate."
We will compare the dose-rate of 0.7475 rad per exposure with the dose-rate of 0.0374 rad per exposure. These tissue-doses are chosen because 0.7475 rad (747.5 millirads) corresponds with an average of 1.0 primary track per nucleus (Table 20-L), and because we want, for comparison, a dose which gives an average of only 0.05 track per nucleus -- which will be a dose 20-fold lower than 0.7475 rad (tracks are proportional to dose) :
(0.7475 rad / nuclear track) x (0.05 track) = 0.0374 rad.
When the track-average = 0.05 , we can see from the percentages above that 95 % of the nuclei are completely escaping irradiation even though the tissue-dose is 0.0374 rad (37.4 millirads) per exposure.
Now let us consider the particular nuclei which do get hit by one or two tracks at the tissue-dose of 37.4 millirads. Each nucleus which is traversed receives the entire energy-transfer in a tiny fraction of a second (Chapter 19).
-
There is no slower transfer of the energy.
These nuclei feel just as much damage as the nuclei which receive one or two tracks at the higher tissue-dose of 0.7475 rads (747.5 millirads). For the nuclei which are hit, the 20-fold reduction in tissue-dose makes no difference. None at all.
Brackenbush and Braby (Brack88, p.252) make a similar point in a discussion of neutrons. We quote: "For a neutron exposure of 3 mGy (30 mSv), only 1.5% of the cell nuclei in the irradiated tissue receive any damage, but they get the same amount as cells exposed to 200 mGy in a dose-response study."
In our comparison of 37.4 versus 747.5 millirads from the 30 KeV X-rays, in spite of the 20-fold reduction in dose-rate (tissue-dose per exposure), 99.9 % of the nuclei exposed at the lower rate have exactly the same experience as most of the nuclei exposed at the higher dose-rate -- namely, either 1.0 track or no track. The only difference between 747.5 millirads per exposure, and 37.4 millirads per exposure, is the fraction of nuclei which experiences two tracks or more.
Further reduction of the dose-rate can only reduce the fraction -- a point now widely recognized (see, for instance, Fein88, p.27). Suppose the 37.4 millirads were delivered evenly over a year. We can regard this as 365 exposures, with each daily exposure about 0.1 millirad. Of course, the same number of primary electron tracks are required to deliver the total dose of 37.4 millirads, regardless of delivery-rate. However, we have just reduced the average frequency of tracks per exposure by a factor of 365. If the Poisson equation were applied to this reduced average, it would show the number of nuclei which ever experience two tracks simultaneously to be tiny, indeed.
Slow Delivery and the Repair-System :
In the region of very low doses, the change in this fraction is the only meaning of "lower dose-rate." Is there any biological meaning to changing this fraction? Let us consider the repair-system. The challenge on the repair-system, within a nucleus, can be reduced for a very few nuclei from two simultaneous tracks to one track, and for the overwhelming percentage of nuclei, the challenge to the repair system cannot be reduced by a lower dose-rate at all. Not at all.
Barendsen, commenting on a report of Hill, Han, and Elkind (Hi84), has also felt compelled to point this out: "Before analysing the interpretations of the dose rate effect suggested by the authors, it should be pointed out that at extremely low doses no difference can possibly exist between high and low dose rates, because at doses where the probability of more than one ionizing particle passing through a cell nucleus is vanishingly small, effects can only be caused by single particle tracks. The concept of dose rate loses its meaning at these very low doses because it depends only on the time in which a single particle traverses a cell" (Bar85).