# Basic Physics of Nuclear Medicine/Attenuation of Gamma-Rays

!The Compton
effect{width="150"}

We covered the interaction of gamma-rays with matter from a descriptive
viewpoint in the previous
chapter and we saw
that the Compton and Photoelectric Effects were the major mechanisms. We
will consider the subject again here but this time from an analytical
perspective. This will allow us to develop a more general understanding
of the phenomenon.

Note that the treatment here also refers to the attenuation of X-rays
since, as we noted before gamma-rays and X-rays are essentially the same
physical entities.

Our treatment begins with a description of a simple radiation experiment
which can be performed easily in the laboratory and which many of the
early pioneers in this field did. We will then build on the information
obtained from such an experiment to develop a simple equation and some
simple concepts which will allow us generalise the situation to any
attenuation situation.

## Attenuation Experiment

The experiment is quite simple. It involves firing a narrow beam of
gamma-rays at a material and measuring how much of the radiation gets
through. We can vary the energy of the gamma-rays we use and the type of
absorbing material as well as its thickness and density.

The experimental set-up is illustrated in the figure below. We refer to
the intensity of the radiation which strikes the absorber as the
**incident intensity**, *I*~0~, and the intensity of the radiation which
gets through the absorber as the **transmitted intensity**, *I*~*x*~.
Notice also that the thickness of the absorber is denoted by *x*.

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From what we covered in the previous
chapter we can
appreciate that some of the gamma-rays will be subjected to interactions
such as the Photoelectric Effect and the Compton Effect as they pass
through the absorber. The transmitted gamma-rays will in the main be
those which pass through without any interactions at all.

We can therefore expect to find that the transmitted intensity will be
less than the incident intensity, that is

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```
$I_x < I_0\,\!$

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But by how much you might ask. Before we consider this let us denote the
difference between *I*~*x*~ and *I*~0~ as ∆*I*, that is

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```
$\Delta I = I_0 - I_x\,\!$

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#### Effect of Atomic Number

:   Let us start exploring the magnitude of ∆*I* by placing different
    absorbers in turn in the radiation beam. What we would find is that
    the magnitude of ∆*I* is highly dependent on the atomic number of
    the absorbing material. For example we would find that ∆*I* would be
    quite low in the case of an absorber made from
    carbon (*Z*=6) and very large in the case of
    lead (*Z*=82).

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:   We can gain an appreciation of why this is so from the following
    figure:

![](NM7_5.gif "NM7_5.gif"){width="220"}

:   The figure illustrates a high atomic number absorber by the large
    circles which represent individual atoms and a low atomic number
    material by smaller circles. The incident radiation beam is
    represented by the arrows entering each absorber from the left.
    Notice that the atoms of the high atomic number absorber present
    larger targets for the radiation to strike and hence the chances for
    interactions via the Photoelectric and Compton Effects is relatively
    high. The attenuation should therefore be relatively large.

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:   In the case of the low atomic number absorber however the individual
    atoms are smaller and hence the chances of interactions are reduced.
    In other words the radiation has a greater probability of being
    transmitted through the absorber and the attenuation is consequently
    lower than in the high atomic number case.

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:   With respect to our spaceship analogy used in the previous
    chapter the
    atomic number can be thought of as the size of individual meteors in
    the meteor cloud.

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:   If we were to precisely control our experimental set-up and
    carefully analyse our results we would find that:\

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```
$\Delta I \propto Z^3\,\!$

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:   Therefore if we were to double the atomic number of our absorber we
    would increase the attenuation by a factor of two cubed, that is 8,
    if we were to triple the atomic number we would increase the
    attenuation by a factor of 27, that is three cubed, and so on.

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:   It is for this reason that high atomic number materials (e.g. Pb)
    are used for radiation protection.

#### Effect of Density

:   A second approach to exploring the magnitude of ∆I is to see what
    happens when we change the density of the absorber. We can see from
    the following figure that a low density absorber will give rise to
    less attenuation than a high density absorber since the chances of
    an interaction between the radiation and the atoms of the absorber
    are relatively lower. In addition, the density determines the
    transmission coefficient as it relates to the sample, since the
    lower the density, the higher the transmission coefficient due to
    the porous nature of the material.

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![](NM7_6.gif "NM7_6.gif"){width="220"}

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:   So in our analogy of the spaceship entering a meteor cloud think of
    meteor clouds of different density and the chances of the spaceship
    colliding with a meteor.

#### Effect of Thickness

:   A third factor which we could vary is the thickness of the absorber.
    As you should be able to predict at this stage the thicker the
    absorber the greater the attenuation.

#### Effect of Gamma-Ray Energy

:   Finally in our experiment we could vary the energy of the gamma-ray
    beam. We would find without going into it in any great detail that
    the greater the energy of the gamma-rays the less the attenuation.
    You might like to think of it in terms of the energy with which the
    spaceship approaches the meteor cloud and the likelihood of a slow
    spaceship getting through as opposed to a spaceship travelling with
    a higher energy.

## Mathematical Model

We will consider a mathematical model here which will help us to express
our experimental observations in more general terms. You will find that
the mathematical approach adopted and the result obtained is quite
similar to what we encountered earlier with Radioactive
Decay. So you will not have to
plod your way through any new maths below, just a different application
of the same form of mathematical analysis!

Let us start quite simply and assume that we vary only the thickness of
the absorber. In other words we use an absorber of the same material
(i.e. same atomic number) and the same density and use gamma-rays of the
same energy for the experiment. Only the thickness of the absorber is
changed.

From our reasoning above it is easy to appreciate that the magnitude of
∆*I* should be dependent on the radiation intensity as well as the
thickness of the absorber, that is for an infinitesimally small change
in absorber thickness:

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```
$-d I \propto I \cdot dx\,\!$

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the minus sign indicating that the intensity is reduced by the absorber.

Turning the proportionality in this equation into an equality, we can
write:

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```
$-d I = \mu I \cdot dx\,\!$

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where the constant of proportionality, μ, is called the Linear
Attenuation Coefficient.

Dividing across by *I* we can rewrite this equation as:

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```
$-\frac{d I}{I} = \mu \cdot dx$

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So this equation describes the situation for any tiny change in absorber
thickness, *dx*. To find out what happens for the complete thickness of
an absorber we simply add up what happens in each small thickness. In
other words we integrate the above equation. Expressing this more
formally we can say that for thicknesses from *x* = 0 to any other
thickness *x*, the radiation intensity will decrease from *I*~0~ to
*I*~*x*~, so that:

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```
$- \int_{I_0}^{I_x} \frac{dI}{I} = \mu \int_{0}^x dx$

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$\therefore \ln \left ( \frac{I_x}{I_0} \right ) = -\mu x$

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$\therefore \frac{I_x}{I_0} = \text{exp}\ (-\mu x)$

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$\therefore I_x = I_0\ \text{exp}\ (-\mu x)\,\!$

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This final expression tells us that the radiation intensity will
decrease in an exponential fashion with the thickness of the absorber
with the rate of decrease being controlled by the Linear Attenuation
Coefficient. The expression is shown in graphical form below. The graph
plots the intensity against thickness, *x*. We can see that the
intensity decreases from *I*~0~, that is the number at *x* = 0, in a
rapid fashion initially and then more slowly in the classic exponential
manner.

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  ------------------------------------------- -------------------------------------------
  ![](NM7_14.gif "NM7_14.gif"){width="220"}   ![](NM7_15.gif "NM7_15.gif"){width="220"}
  ------------------------------------------- -------------------------------------------

  : Graphical representation of the dependence of radiation intensity on
  the thickness of absorber: Intensity versus thickness on the left and
  the natural logarithm of the intensity versus thickness on the right.

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The influence of the Linear Attenuation Coefficient can be seen in the
next figure. All three curves here are exponential in nature, only the
Linear Attenuation Coefficient is different. Notice that when the Linear
Attenuation Coefficient has a low value the curve decreases relatively
slowly and when the Linear Attenuation Coefficient is large the curve
decreases very quickly.

!Exponential attenuation expressed using a small, medium and large
value of the Linear Attenuation Coefficient,
µ.

The Linear Attenuation Coefficient is characteristic of individual
absorbing materials. Some like carbon have a small value and are easily
penetrated by gamma-rays. Other materials such as lead have a relatively
large Linear Attenuation Coefficient and are relatively good absorbers
of radiation:

  Absorber        100 keV    200 keV    500 keV
  --------------- ---------- ---------- ----------
  **Air**         0.000195   0.000159   0.000112
  **Water**       0.167      0.136      0.097
  **Carbon**      0.335      0.274      0.196
  **Aluminium**   0.435      0.324      0.227
  **Iron**        2.72       1.09       0.655
  **Copper**      3.8        1.309      0.73
  **Lead**        59.7       10.15      1.64

  : Linear Attenuation Coefficients (in cm^-1^) for a range of materials
  at gamma-ray energies of 100, 200 and 500 keV.

The materials listed in the table above are air, water and a range of
elements from carbon (*Z*=6) through to lead (*Z*=82) and their Linear
Attenuation Coefficients are given for three gamma-ray energies. The
first point to note is that the Linear Attenuation Coefficient increases
as the atomic number of the absorber increases. For example it increases
from a very small value of 0.000195 cm^-1^ for air at 100 keV to almost
60 cm^-1^ for lead. The second point to note is that the Linear
Attenuation Coefficient for all materials decreases with the energy of
the gamma-rays. For example the value for copper decreases from about
3.8 cm^-1^ at 100 keV to 0.73 cm^-1^ at 500 keV. The third point to note
is that the trends in the table are consistent with the analysis
presented earlier.

Finally it is important to appreciate that our analysis above is only
strictly true when we are dealing with narrow radiation beams. Other
factors need to be taken into account when broad radiation beams are
involved.

## Half Value Layer

As with using the Half Life to describe the Radioactive Decay
Law an indicator is usually
derived from the exponential attenuation equation above which helps us
think more clearly about what is going on. This indicator is called the
**Half Value Layer** and it expresses the thickness of absorbing
material which is needed to reduce the incident radiation intensity by a
factor of two. From a graphical point of view we can say that when:

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$I_x = \frac{I_0}{2}$

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the thickness of absorber is the Half Value Layer:

![](NM7_19.gif "NM7_19.gif"){width="320"}

The Half Value Layer for a range of absorbers is listed in the following
table for three gamma-ray energies:

  Absorber        100 keV   200 keV   500 keV
  --------------- --------- --------- ---------
  **Air**         3555      4359      6189
  **Water**       4.15      5.1       7.15
  **Carbon**      2.07      2.53      3.54
  **Aluminium**   1.59      2.14      3.05
  **Iron**        0.26      0.64      1.06
  **Copper**      0.18      0.53      0.95
  **Lead**        0.012     0.068     0.42

  : Half Value Layers (in cm) for a range of materials at gamma-ray
  energies of 100, 200 and 500 keV.

The first point to note is that the Half Value Layer decreases as the
atomic number increases. For example the value for air at 100 keV is
about 35 meters and it decreases to just 0.12 mm for lead at this
energy. In other words 35 m of air is needed to reduce the intensity of
a 100 keV gamma-ray beam by a factor of two whereas just 0.12 mm of lead
can do the same thing. The second thing to note is that the Half Value
Layer increases with increasing gamma-ray energy. For example from 0.18
cm for copper at 100 keV to about 1 cm at 500 keV. Thirdly note that
relative to the data in the previous table there is a reciprocal
relationship between the Half Value Layer and the Linear Attenuation
Coefficient, which we will now investigate.

## Relationship between μ and the HVL

As was the case with the Radioactive Decay Law, where we explored the
relationship between the Half Life and the Decay Constant, a
relationship can be derived between the Half Value Layer and the Linear
Attenuation Coefficient. We can do this by using the definition of the
Half Value Layer:

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$I = \frac{I_0}{2}$

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when

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$x = x_{\frac{1}{2}}$

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and inserting it in the exponential attenuation equation, that is:

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$I = I_0\ \text{exp}\ (-\mu x)\,\!$

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to give

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$\frac{I_0}{2} = I_0\ \text{exp}\ (-\mu x_{\frac{1}{2}})$

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Therefore

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$\frac{1}{2} = \text{exp}\ (-\mu x_{\frac{1}{2}})$

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$\therefore 2^{-1} = \text{exp}\ (-\mu x_{\frac{1}{2}})$

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$\therefore \ln 2^{-1} = -\mu x_{\frac{1}{2}}$

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$\therefore \ln 2 = \mu x_{\frac{1}{2}}$

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$\therefore 0.693 = \mu x_{\frac{1}{2}}$

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$\mu = \frac{0.693}{x_{\frac{1}{2}}}$

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and

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$x_{\frac{1}{2}} = \frac{0.693}{\mu}$

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These last two equations express the relationship between the Linear
Attenuation Coefficient and the Half Value Layer. They are very useful
as you will see when solving numerical questions relating to attenuation
and frequently form the first step in solving a numerical problem.

## Mass Attenuation Coefficient

We implied above that the Linear Attenuation Coefficient was useful when
we were considering an absorbing material of the same density but of
different thicknesses. A related coefficient can be of value when we
wish to include the density, ρ, of the absorber in our analysis. This is
the **Mass Attenuation Coefficient** which is defined as the:

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$\frac{\text{Linear Attenuation Coefficient}}{\text{Density}} = \frac{\mu}{\rho}$

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The measurement unit used for the Linear Attenuation Coefficient in the
table above is cm^-1^, and a common unit of density is the g cm^-3^. You
might like to derive for yourself on this basis that the cm^2^ g^-1^ is
the equivalent unit of the Mass Attenuation Coefficient.

## Questions

Two questions are given below to help you develop your understanding of
the material presented in this chapter. The first one is relatively
straight-forward and will exercise your application of the exponential
attenuation equation. The second question is a lot more challenging and
will help you relate exponential attenuation to radioactivity and
radiation exposure.

**Question 1**

How much aluminium is required to reduce the intensity of a 200 keV
gamma-ray beam to 10% of its incident intensity? Assume that the Half
Value Layer for 200 keV gamma-rays in Al is 2.14 cm.

**Answer**

:   The question phrased in terms of the symbols used above is:

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$I = \frac{I_0}{10},\ \text{when}\ x = \text{?}$

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:   We are told that the Half Value Layer is 2.14 cm. Therefore the
    Linear Attenuation Coefficient is

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$\mu = \frac{0.693}{x_{\frac{1}{2}}} = \frac{0.693}{2.14} = 0.324\ \text{cm}^{-1}$

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:   Now combining all this with the exponential attenuation equation:

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$I = I_0\ \text{exp}\ (-\mu x)\,\!$

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:   we can write:

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$\frac{I_0}{10} = I_0\ \text{exp}\ (-0.324x)$

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:   Therefore

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$\frac{1}{10} = \text{exp}\ (-0.324x)$

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$\therefore -\ln 10 = -0.324x$

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$\therefore x = \frac{\ln 10}{0.324} = \frac{2.3}{0.324} = 7.1\ \text{cm}$

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$\therefore x \approx 7\ \text{cm}$

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:   So the thickness of aluminium required to reduce these gamma-rays by
    a factor of ten is about 7 cm. This relatively large thickness is
    the reason why aluminium is not generally used in radiation
    protection - its atomic number is not high enough for efficient and
    significant attenuation of gamma-rays.

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:   You might like to try this question for the case when Pb is the
    absorber - but you will need to find out the Half Value Layer for
    the 200 keV gamma-rays yourself!

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:   Here\'s a hint though: have a look at one of the tables above.

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:   And here\'s the answer for you to check when you\'ve finished:
    `<b>`{=html}2.2 mm`</b>`{=html}.

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:   In other words a relatively thin thickness of Pb is required to do
    the same job as 7 cm of aluminium.

**Question 2**

A 10^5^ MBq source of ^137^Cs is to be contained in a Pb box so that the
exposure rate 1 m away from the source is less than 0.5 mR/hour. If the
Half Value Layer for ^137^Cs gamma-rays in Pb is 0.6 cm, what thickness
of Pb is required? The Specific Gamma Ray Constant for ^137^Cs is 3.3 R
hr^-1^ mCi^-1^ at 1 cm.

**Answer**

:   This is a fairly typical question which arises when someone is using
    radioactive materials. We wish to use a certain quantity of the
    material and we wish to store it in a lead container so that the
    exposure rate when we are working a certain distance away is below
    some level for safety reasons. We know the radioactivity of the
    material we will be using. But its quoted in SI units. We look up a
    reference book to find out the exposure rate for this radioisotope
    and find that the Specific Gamma Ray Constant is quoted in
    traditional units. Just as in our question!

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:   So let us start by getting our units right. The Specific Gamma Ray
    Constant is given as:

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3.3 R hr^-1^ mCi^-1^ at 1 cm from the source.

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:   This is equal to:

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3300 mR hr^-1^ mCi^-1^ at 1 cm from the source,

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:   which is equal to:

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$\frac{3300}{(100)^2}\ \text{mR hr}^{-1}\ \text{mCi}^{-1}\ \text{at 1 m from the source}$

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:   on the basis of the Inverse Square Law. This result expressed per
    becquerel is

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$\frac{3300}{10^4 (3.7 \cdot 10^7)}\ \text{mR hr}^{-1}\ \text{Bq}^{-1}\ \text{at 1 m from the source}$

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:   since 1 mCi = 3.7 x 10^7^ Bq. And therefore for 10^5^ MBq, the
    exposure rate is:

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$\frac{3300 \cdot 10^5 \cdot 10^6}{10^4 (3.7 \cdot 10^7)}\ \text{mR hr}^{-1}\ (10^5\ \text{MBq})^{-1}\ \text{at 1 m from the source}$

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:   That is the exposure rate 1 meter from our source is
    `<b>`{=html}891.9 mR hr^-1^`</b>`{=html}.

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:   We wish to reduce this exposure rate according to the question to
    less than 0.5 mR hr^-1^ using Pb.

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:   You should be able at this stage to use the exponential attenuation
    equation along with the Half Value Layer for these gamma-rays in Pb
    to calculate that the thickness of Pb required is about
    `<b>`{=html}6.5 cm`</b>`{=html}.

## External Links

-   Mucal on the Web - an online
    program which calculates x-ray absorption coefficients - by
    Pathikrit Bandyopadhyay, The Center for Synchrotron Radiation
    Research and Instrumentation at the Illinois Institute of
    Technology.
-   Tables of X-Ray Mass Attenuation
    Coefficients -
    a vast amount of data for all elements from National Institute of
    Science & Technology, USA.

