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CHAPTER 20
Number of Primary Electron-Tracks per Cell-Nucleus, per Rad of Dose
Received from Various Sources of Radiation
----------------------------------------------------------------------------
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This chapter is arranged in three parts:
1. List of the Information Needed to Calculate the Number of Tracks per
Nucleus, p.1
2. Provision of the Input, the Calculations, and the Answer, p.2
3. 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!}
Where:
* -- 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).
------------------------------------------------------------------
============================================================================
----------------------------------------------------------------------------
Table 20-C
Photons Required to Deliver a Dose of 1 Centi-Gray ( 1 Rad ) to 1 Gram of
Tissue.
-------------------------------------------------------------
(Item A): 6.24 x 10^10 KeV required to deliver 1 centi-gray (cGy) to 1 gram
of tissue.
================================================================================
| Item B | (Item A / Item B) | Item C |
| | | |
| INITIAL | ENERGY REQUIRED, | NUMBER of PHOTONS |
| PHOTON | DIVIDED BY | REQUIRED TO DELIVER |
|PHOTON SOURCE ENERGY | PHOTON ENERGY | 1 RAD TO 1 GRAM |
| | | |
|==============================================================================|
| | | |
|Medical X-rays 30 KeV | 6.24 x 10^10 / 30 | 2.08E+09 |
| | | or 2.08 x 10^9 photons |
| | | or 2.08 billion photons |
| | | or 2,080,000,000 photons|
| | | |
|Radium-226 and | | |
|daughters 596 KeV | 6.24 x 10^10 / 596 | 1.05E+08 |
| | | or 1.05 x 10^8 photons |
| | | or 105 million photons |
| | | or 105,000,000 photons |
| | | |
|Cesium-137 662 KeV | 6.24 x 10^10 / 662 | 9.43E+07 |
| | | or 9.43 x 10^7 photons |
| | | or 94.3 million photons |
| | | or 94,300,000 photons |
| | | |
|A-Bomb Gammas 1608 KeV | 6.24 x 10^10 / 1608 | 3.88E+07 |
| | | or 3.88 x 10^7 photons |
| | | or 38.8 million photons |
| | | or 38,800,000 photons |
================================================================================
NOTES -----
Item C is the number of photons whose energy must be totally absorbed in one
gram of tissue to produce a tissue-dose of one centi-gray (one rad) in that
gram.
For readers who are not yet familiar with the exponential format for
numbers, we have used the space available in this table to express these
rather large numbers in additional ways.
----------------------------------------------------------------------------
============================================================================
============================================================================
----------------------------------------------------------------------------
Table 20-D
Energies of Electrons in the "Packets" Associated with Various Photons.
-------------------------------------------------------------
Data are brought forward from Tables 32-A, 32-B, and 32-C.
(Energies are given in both KeV and MeV)
================================================================
| Item D |
|PHOTON SOURCE TYPE OF ELECTRON ELECTRON |
| ELECTRON ENERGY ENERGY |
| (Kev) (MeV) |
|==============================================================|
|30 KeV Photons |
|Medical X-rays |
| Photo-electron 30 0.030 |
|==============================================================|
|596 KeV Photons |
|Radium-226 and |
|daughters |
| Compton 208.583 0.2086 |
| Compton 116.728 0.1167 |
| Compton 69.626 0.0696 |
| Compton 44.273 0.0443 |
| Compton 29.813 0.0298 |
| Compton 21.078 0.0211 |
| Compton 15.516 0.0155 |
| Photo-electron 90.384 0.0904 |
|==============================================================|
|662 KeV Photons |
|Cesium-137 |
| Compton 238.825 0.2388 |
| Compton 131.931 0.1319 |
| Compton 77.571 0.0776 |
| Compton 48.656 0.0487 |
| Compton 32.377 0.0324 |
| Compton 22.663 0.0227 |
| Compton 16.546 0.0165 |
| Photo-electron 93.429 0.0934 |
|==============================================================|
|1608 KeV Photons |
|A-Bomb Gammas |
| Compton 693.765 0.6938 |
| Compton 357.271 0.3573 |
| Compton 190.906 0.1909 |
| Compton 107.792 0.1078 |
| Compton 64.914 0.0649 |
| Compton 41.645 0.0416 |
| Compton 28.259 0.0283 |
| Compton 20.107 0.0201 |
| Compton 14.880 0.0149 |
| Photo-electron 88.460 0.0885 |
================================================================
----------------------------------------------------------------------------
These tabulations are based on the assumption that all of the energy of the
original photon is absorbed in the tissue, for those photons which interact
at all with the tissue. See text (C) and (D).
----------------------------------------------------------------------------
============================================================================
============================================================================
----------------------------------------------------------------------------
Table 20-FG
Ranges and Cell-Traversals for Various Primary Electrons. (Micrometer =
Micron.)
==============================================================================================
| Item D Item F Item E Item G | Item G (sum) |
| ELECTRON ELECTRON CUBOIDAL NUMBER | CELLS TRAVERSED PER|
|PHOTON SOURCE ENERGY RANGE CELL DEPTH OF CELLS | ELECTRON, OR PER |
| (KeV) (microns) (microns) TRAVERSED | ELECTRON-PACKET |
|============================================================================================|
|30 KeV Photons | |
|Medical X-rays 30 19.779 11.4 1.735 | 1.735 |
|============================================================================================|
| | |
|596 KeV Photons 208.582 462.635 11.4 40.582 | |
|Radium-226 + daughters 116.727 150.107 11.4 13.167 | |
| 69.626 73.866 11.4 6.479 | |
| 44.273 38.353 11.4 3.364 | |
| 29.813 19.543 11.4 1.714 | |
| 21.078 8.687 11.4 0.762 | |
| 15.516 1.975 11.4 0.173 | |
| 90.384 105.343 11.4 9.241 | |
| Sum = | 75.483 |
|============================================================================================|
| | |
|662 KeV Photons 238.825 566.577 11.4 49.700 | |
|Cesium-137 131.931 185.777 11.4 16.296 | |
| 77.571 85.659 11.4 7.514 | |
| 48.656 44.262 11.4 3.883 | |
| 32.377 22.802 11.4 2.000 | |
| 22.663 10.629 11.4 0.932 | |
| 16.546 3.206 11.4 0.281 | |
| 93.429 110.143 11.4 9.662 | |
| Sum = | 90.268 |
|============================================================================================|
| | |
|1608 KeV Photons 693.765 2517.482 11.4 220.832 | |
|A-Bomb Gammas 357.271 1031.762 11.4 90.505 | |
| 190.906 384.423 11.4 33.721 | |
| 107.792 132.116 11.4 11.589 | |
| 64.914 67.022 11.4 5.879 | |
| 41.645 34.857 11.4 3.058 | |
| 28.259 17.583 11.4 1.542 | |
| 20.107 7.504 11.4 0.658 | |
| 14.880 1.217 11.4 0.107 | |
| 88.460 102.335 11.4 8.977 | |
| Sum = | 376.869 |
==============================================================================================
Item D entries come from Table 20-D.
Item F entries are brought forward from Chapter 33. For 30 KeV photons,
there are only photo-electrons, all of the same energy. For all other
classes, all entries except the lowest row are for Compton electrons. The
lowest row is for photo-electrons.
Item E entry, depth of a cuboidal cell, is from the text. Micron =
Micrometer.
Item G entries, number of cell-traversals per electron, are calculated:
(Range of the Electron / Depth of Cell), or (Item F / Item E).
----------------------------------------------------------------------------
============================================================================
============================================================================
----------------------------------------------------------------------------
Table 20-H
Total Cell-Traversals for 1 Rad (cGy) of Dose Delivered to 1 Gram of
Cellular Tissue.
==================================================================================
| Item C Item G (sum) Item H |
| |
| Origin of Packets Required Cell-Traversals by Total Cell-Traversals |
| Electron to Deliver Electrons of a in Delivery of 1 Rad |
| Packets One Rad (cGy) Single Packet to 1 Gram |
|================================================================================|
|30 KeV X-Rays 2.08E+09 1.735 3.61E+09 or 3.61 billion|
| |
|596 KeV gammas 1.05E+08 75.483 7.93E+09 or 7.93 billion|
|Source: Radium-226 |
| |
|662 KeV gammas 9.43E+07 90.268 8.51E+09 or 8.51 billion|
|Source: Cesium-137 |
| |
|1608 KeV gammas 3.88E+07 376.869 1.46E+10 or 14.6 billion|
|Source: A-bomb |
==================================================================================
Item C entries are from Table 20-C.
Item G entries are from Table 20-FG.
Item H entries are calculated: (Packets x Cell-Traversals per Packet),
or (Item C x Item G, sum).
"Cell-traversals" does NOT mean separate cells traversed. Some cells
experience multiple traversals.
----------------------------------------------------------------------------
============================================================================
============================================================================
----------------------------------------------------------------------------
Table 20-(Eye)
Total Nuclear-Traversals for 1 Rad (cGy) of Dose Delivered to 1 Gram of
Cellular Tissue.
=====================================================================================
| Item H Item I (Eye) |
| |
| Origin of Total Cell Traversals Total Nuclear-Traversals by Primary Tracks|
| Electron in Delivery of in Delivery of |
| Packets One cGy (Rad) 1 cGy (Rad) to 1 Gram |
|===================================================================================|
|30 KeV X-Rays 3.61E+09 9.03E+08 or 903 million |
| |
|596 KeV gammas 7.93E+09 1.98E+09 or 1.98 billion |
|Source: Radium-226 |
| |
|662 KeV gammas 8.51E+09 2.13E+09 or 2.13 billion |
|Source: Cesium-137 |
| |
|1608 KeV gammas 1.46E+10 3.65E+09 or 3.65 billion |
|Source: A-bomb |
=====================================================================================
Item I (Eye) entries are (Item H) x (0.25). This follows from the model in
which cells and cell-nuclei are cuboidal, and electrons come in normal to
one face. See text, parts (E) and (I, or Eye).
----------------------------------------------------------------------------
============================================================================
============================================================================
----------------------------------------------------------------------------
Table 20-K
Average Number of Primary Electron Tracks per Nucleus at a Dose of 1 Rad (1
cGy).
===========================================================================
| Item I (Eye) Item J Item K |
| |
|ORIGIN OF TOTAL NUCLEAR- NUMBER OF AVERAGE NUMBER OF |
|ELECTRON TRAVERSALS NUCLEI NUCLEAR TRAVERSALS |
|PACKETS FOR DELIVERY PER GRAM PER NUCLEUS, AT A |
| OF 1 cGy OF TISSUE TISSUE-DOSE OF 1 cGy|
|=========================================================================|
|30 KeV X-rays 9.03E+08 6.75E+08 1.3378 |
| |
|596 KeV gammas |
|Source: Radium-226 1.98E+09 6.75E+08 2.9370 |
| |
|662 KeV gammas |
|Source: Cesium-137 2.13E+09 6.75E+08 3.1556 |
| |
|1608 KeV gammas |
|Source: A-bomb 3.65E+09 6.75E+08 5.4074 |
===========================================================================
----------------------------------------------------------------------------
Item J entry comes from the text, part (J).
Item K entries are calculated: (Total nuclear-traversals / number of
nuclei available to be traversed), or (Item I / Item J). The "average number
of nuclear traversals per nucleus at 1 rad" is the same as "average number
of primary ionization tracks per nucleus at one rad."
It will be exceedingly rare for a nucleus to be traversed by two
high-speed electrons originating from the SAME initial photon. Each time the
Compton process occurs, it creates only one high speed electron plus a new
photon of reduced energy. The new photon, traveling at the speed of light,
is in a different location when the next Compton electron is produced.
----------------------------------------------------------------------------
============================================================================
============================================================================
----------------------------------------------------------------------------
Table 20-L
Tissue-Dose in Centi-Gray When the Average Track-Rate per Nucleus Is ONE.
==================================================================================
| Item K Item L |
|ORIGIN OF AVERAGE NUCLEAR-TRAVERSALS DOSE IN CENTI-GRAY WHEN AVERAGE |
|ELECTRON PER NUCLEUS IN DELIVERY TRACK-RATE PER NUCLEUS |
|PACKETS OF 1 cGy ( 1 RAD ) IS ONE |
| [ Tracks / Nucleus / cGy ] [ cGy / Track / Nucleus ] |
|================================================================================|
|30 KeV X-rays 1.3378 0.7475 cGy, or 747.5 millirads |
| |
|596 KeV gammas |
|Source: Radium-226 2.937 0.3405 cGy, or 340.5 millirads |
| |
|662 KeV gammas |
|Source: Cesium-137 3.156 0.3169 cGy, or 316.9 millirads |
| |
|1608 KeV gammas |
|Source: A-bomb 5.4074 0.18493 cGy, or 184.9 millirads |
==================================================================================
Item L entries are calculated: ( 1 / Item K).
----------------------------------------------------------------------------
============================================================================
============================================================================
----------------------------------------------------------------------------
Table 20-M
"If ... Then" : Relationship between Tissue-Doses and Average Tracks per
Nucleus.
Tracks refer to primary ionization tracks, or primary electron tracks.
===================================================================================================
| For 30 KeV X-Rays || For 596 KeV Gamma Rays (Ra-226) |
|----------------------------------------------||-------------------------------------------------|
| If Total Then the | If Average Then the || If Total Then the | If Average Then the|
|Tissue-Dose Average | Number of Tissue- || Tissue-Dose Average | Number of Tissue-|
| Is This Number of | Tracks / Dose || Is This Number of | Tracks / Dose|
| (in Rads) Tracks / | Nucleus in Rads || (in Rads) Tracks / | Nucleus in Rads|
| (or cGy) Nucleus | Is (or cGy) || (or cGy) Nucleus | Is (or cGy)|
| Is | Is || Is | Is|
|----------------------------------------------||-------------------------------------------------|
| 0.001 0.0013 | 1 0.7475 || 0.001 0.0029 | 1 0.3405|
| 0.010 0.0134 | 2 1.4950 || 0.010 0.0294 | 2 0.6810|
| 0.050 0.0669 | 3 2.2425 || 0.050 0.1468 | 3 1.0215|
| 0.100 0.1338 | 4 2.9900 || 0.100 0.2937 | 4 1.3620|
| 0.500 0.6689 | 5 3.7375 || 0.3405 1.0000 | 5 1.7025|
| 0.7475 1.0000 | 6 4.4850 || 0.500 1.4684 | 6 2.0430|
| 1.000 1.3378 | 7 5.2325 || 1.000 2.9370 | 7 2.3835|
| 5.000 6.6890 | 8 5.9800 || 5.000 14.6843 | 8 2.7240|
| 10.000 13.3779 | 9 6.7275 || 10.000 29.3686 | 9 3.0645|
| 50.000 66.8896 | 10 7.4750 || 50.000 146.8429 | 10 3.4050|
| 100.000 133.7793 | 20 14.9500 || 100.000 293.6858 | 20 6.8100|
| 200.000 267.5585 | 50 37.3750 || 200.000 587.3715 | 50 17.0250|
| | 100 74.7500 || | 100 34.0500|
|=================================================================================================|
| For 662 KeV Gamma Rays (Cs-137) || For 1608 KeV Gamma Rays (A-Bomb) |
|----------------------------------------------||-------------------------------------------------|
| If Total Then the | If Average Then the || If Total Then the | If Average Then the|
|Tissue-Dose Average | Number of Tissue- || Tissue-Dose Average | Number of Tissue-|
| Is This Number of | Tracks / Dose || Is This Number of | Tracks / Dose|
| (in Rads) Tracks / | Nucleus in Rads || (in Rads) Tracks / | Nucleus in Rads|
| (or cGy) Nucleus | Is (or cGy) || (or cGy) Nucleus | Is (or cGy)|
| Is | Is || Is | Is|
|----------------------------------------------||-------------------------------------------------|
| 0.001 0.0032 | 1 0.3169 || 0.001 0.0054 | 1 0.1849|
| 0.010 0.0316 | 2 0.6338 || 0.010 0.0541 | 2 0.3698|
| 0.050 0.1578 | 3 0.9507 || 0.050 0.2704 | 3 0.5547|
| 0.100 0.3156 | 4 1.2676 || 0.100 0.5407 | 4 0.7396|
| 0.3169 1.0000 | 5 1.5845 || 0.1849 1.0000 | 5 0.9245|
| 0.500 1.5778 | 6 1.9014 || 0.500 2.7042 | 6 1.1094|
| 1.000 3.1556 | 7 2.2183 || 1.000 5.4074 | 7 1.2943|
| 5.000 15.7778 | 8 2.5352 || 5.000 27.0416 | 8 1.4792|
| 10.000 31.5557 | 9 2.8521 || 10.000 54.0833 | 9 1.6641|
| 50.000 157.7785 | 10 3.1690 || 50.000 270.4164 | 10 1.8490|
| 100.000 315.5570 | 20 6.3380 || 100.000 540.8329 | 20 3.6980|
| 200.000 631.1139 | 50 15.8450 || 200.000 1081.666 | 50 9.2450|
| | 100 31.6900 || | 100 18.4900|
===================================================================================================
----------------------------------------------------------------------------
============================================================================
============================================================================
----------------------------------------------------------------------------
Table 20-N
Poisson Distribution of Events (Nuclear Tracks). Equation Is in the Text,
(Part N).
Event means a primary ionization track somewhere through a cell-nucleus.
----------------------------------------------------------------------------
Probability (or Chance) of Exactly V Primary Tracks per Nucleus When Average
Number of Tracks (N) Equals 1.0
Mean number of tracks = 1.0 track per nucleus.
"Question: How many nuclei will receive exactly 0, 1, 2, 3, 4, 5, etc.
tracks?"
_________________________________________________________________________
0.367879 chance of zero events if average = 1.0
0.367879 chance of 1 event if average = 1.0
0.183940 chance of 2 events if average = 1.0
0.061313 chance of 3 events if average = 1.0
0.015328 chance of 4 events if average = 1.0
0.003066 chance of 5 events if average = 1.0
0.000511 chance of 6 events if average = 1.0
0.000073 chance of 7 events if average = 1.0
0.000009 chance of 8 events if average = 1.0
0.000001 chance of 9 events if average = 1.0
The sum of probabilities must add up to 1.0000. By the time we reach 9
events per nucleus, the sum is exceedingly close to 1.0000: Sum of
probabilities of zero or more events = 0.999999
Sum of probabilities of 2 or more events when average is one: 0.264241 (26.4
per 100 nuclei).
----------------------------------------------------------------------------
Probability (or Chance) of Exactly V Primary Tracks per Nucleus When Average
Number of Tracks (N) Equals 0.05
Mean number of tracks = 0.05 track per nucleus.
"Question: How many nuclei will receive exactly 0, 1, 2, 3, etc. tracks?"
___________________________________________________________________________
Probability of zero track = 0.951229
Probability of exactly 1 track = 0.047561
Probability of exactly 2 tracks = 0.001189
Probability of 3 or more tracks = 2.1E-05
----------------------------------------------------------------------------
Probability (or Chance) of Exactly V Primary Tracks per Nucleus When Average
Number of Tracks (N) Equals 0.02937
Mean number of tracks = 0.02937 track per nucleus.
"Question: How many nuclei will receive exactly 0, 1, 2, 3, etc. tracks?"
___________________________________________________________________________
Probability of zero track = 0.971057 (97.1 percent)
Probability of exactly 1 track = 0.028519 (2.85 percent)
Probability of exactly 2 tracks = 0.000418 (418 per million nuclei)
Probability of exactly 3 tracks = 0.000004 (4 per million nuclei)
Probability of exactly 4 tracks = 3.01E-08 (3 per 100 million nuclei)
Probability of exactly 5 tracks = 1.77E-10 (2 per 10 billion nuclei)
In Chapter 21, tables also provide Poisson distributions when average is
0.2937 and 0.6689 track.
----------------------------------------------------------------------------
============================================================================
============================================================================
----------------------------------------------------------------------------
Table 20-"O" 83.3 and 100 KeV Photo-Electrons :
Estimated Number of Primary Electron Tracks per Nucleus, at a Dose of 1 Rad
(cGy).
----------------------------------------------------------------------------
Experimental irradiation of cells in the laboratory often involves
X-irradiation from a 250 kilovolt machine -- which means that the average
energy of the photons in the resulting X-ray beam is about one-third of the
maximum 250 KeV. So the average photon energy is about 83.3 KeV.
We have used the approximation that, at 100 KeV or less, the photon's
energy is transferred entirely to a single photo-electron, although the
Compton process will occur sometimes too (Chapter 32 text, and Table 32-A,
Note 6).
In this table, we will derive an estimate of the average number of tracks
per cell-nucleus, at a dose of one rad (cGy) delivered by 83.3 KeV
photo-electrons. We include the corresponding estimate for 100 KeV
photo-electrons too.
----------------------------------------------------------------------------
Photons Required to Deliver a Dose of 1 Rad (6.24 x 10^10 KeV) to 1 Gram of
Tissue (Item C) :
Number of photons required = (Total energy needed) / (Energy per photon).
Total energy needed Energy per photon Photons required for 1 rad to one gram.
6.24 x 10^10 KeV 83.3 KeV 7.49E+08 photons, which all become electrons.
6.24 x 10^10 KeV 100 KeV 6.24E+08 photons, which all become electrons.
----------------------------------------------------------------------------
Estimated Range of 83.3 KeV Photo-Electrons (Item F) :
In Table 20-FG, we have two values between which we can interpolate:
Energy (KeV) Range (microns)
Cs-137 77.571 85.659
Cs-137 93.429 110.143
Difference 15.858 24.484
But we are going from 77.571 to 83.3, or 5.429 KeV.
So we pick up (5.429 / 15.858) of the difference: (5.429 / 15.858) x
(24.484) microns,
which is 8.38212 microns.
Total range for 83.3 KeV photo-electrons = 85.659 microns + 8.382 microns =
94.041 microns.
Estimated Range of 100 KeV Photo-ElectronsS (Item F) :
In Table 20-FG, we have two values between which we can interpolate:
Energy (KeV) Range (microns)
Cs-137 93.429 110.143
A-bomb 107.792 132.116
Difference 14.363 21.973
But we are going from 93.429 to 100, or 6.571 KeV.
So we pick up (6.571 / 14.363) of the difference: (6.571 / 14.363) x
(21.973) microns,
which is 10.05253 microns.
Total range for 100 KeV photo-electrons = 110.143 microns + 10.053 microns =
120.196 microns.
----------------------------------------------------------------------------
Total Cell Traversals in Delivery of 1 Rad to 1 Gram (Item H) :
Cell traversals per electron = (Range per electron in microns) / (11.4
microns per cell -- Item E).
Total cell traversals = (Cell traverals per electron) x (Number of electrons
required for 1 rad).
Photo-electrons Cell traversals per electron Total cell traversals (Item H):
83.3 KeV 94.041 / 11.4 = 8.249 8.249 x 7.49E+08 = 6.18E+09
100 KeV 120.196 / 11.4 = 10.544 10.544 x 6.24E+08 = 6.58E+09
----------------------------------------------------------------------------
Estimated Number of Primary Electron Tracks per Nucleus, at a Dose of 1 Rad (cGy).
Total nuclear traversals in delivery of 1 rad = (Total cell traversals) x (0.25; see Table 20-Eye).
Tracks = (Total nuclear traversals) / (Number of nuclei available, or 6.75E+08 nuclei per gram).
Photo-electrons Total nuclear traversals Nuclear traversals / nuclei available
83.3 KeV (0.25 x 6.18E+09) = 1.54E+09 (1.54E+09 / 6.75E+08) = 2.28 tracks
100 KeV (0.25 x 6.58E+09) = 1.64E+09 (1.64E+09 / 6.75E+08) = 2.43 tracks
----------------------------------------------------------------------------
============================================================================