# Basic Physics of Nuclear Medicine/The Radioactive Decay Law

!Graph of the stability of every known nucleus, plotted as Z (number of
protons) versus N (number of neutrons). The color corresponds to the
value of the half-life T½ with a strong log scale, since it varies
between 10^−20^ and 10^20^
seconds. versus N (number of neutrons). The color corresponds to the value of the half-life T½ with a strong log scale, since it varies between 10−20 and 1020 seconds.")

We covered radioactive decay from a phenomenological perspective in the
last
chapter.
In this chapter we consider the topic from a more general analytical
perspective.

The reason for doing this is so that we can develop a form of thinking
which will help us to understand what is going on in a quantitative,
mathematical sense. We will be introduced to concepts such as the
**Decay Constant** and the **Half Life** as well as units used for the
measurement of radioactivity. You will also have a chance to develop
your understanding by being brought through three questions on this
subject.

## Assumptions

The usual starting point in most forms of analysis in physics is to make
some assumptions which simplify the situation. By simplifying the
situation we can dispose of irrelevant effects which tend to complicate
matters but in doing so we sometimes make the situation so simple that
it becomes a bit too abstract and apparently hard to understand.

For this reason we will try here to relate the subject of radioactive
decay to a more common situation which we will use as an analogy and
hopefully we will be able to overcome the abstract feature of the
subject matter. The analogy we will use here is that of making
popcorn.

So think about putting some oil in a pot, adding the corn, heating the
pot on the cooker and watching what happens. You might also like to try
this out while considering the situation!

For our radioactive decay situation we first of all consider that we
have a sample containing a large number of radioactive nuclei all of the
same kind. This is our unpopped corn in the pot for example.

Secondly we assume that all of the radioactive nuclei decay by the same
process be it alpha, beta or gamma-decay. In other words our unpopped
corn goes pop at some stage during the heating process.

Thirdly take a few moments to ponder on the fact that we can only really
consider what is going on from a statistical perspective. If you look at
an individual piece of corn, can you figure out when it is going to pop?
Not really. You can however figure out that a large number of them will
have popped after a period of time. But its rather more difficult to
figure out the situation for an individual piece of corn. So instead of
dealing with individual entities we consider what happens on a larger
scale and this is where statistics comes in.
We can say that the radioactive decay is a statistical one-shot process,
that is when a nucleus has decayed it cannot repeat the process again.
In other words when a piece of corn has popped it cannot repeat the
process. Simple!

In addition as long as a radioactive nucleus has not decayed the
probability for it doing so in the next moment remains the same. In
other words if a piece of corn has not popped at a certain time the
chance of it popping in the next second is the same as in the previous
second. The bets are even!

Let us not push this popcorn analogy too far though in that we know that
we can control the rate of popping by the heat we apply to the pot for
example. However as far as our radioactive nuclei are concerned there is
nothing we can do to control what is going on. The rate at which nuclei
go pop (or **decay**, in other words) cannot be influenced by heating up
the sample. Nor by cooling it for that matter or by putting it under
greater pressures, by changing the gravitational environment by taking
it out into space for instance, or by changing any other aspect of its
physical environment. The only thing that determines whether an
individual nucleus will decay seems to be the nucleus itself. But on the
average we can say that it will decay at some stage.

## The Radioactive Decay Law

Let us now use some symbols to reduce the amount of writing we have to
do to describe what is going on and to avail ourselves of some
mathematical techniques to simplify the situation even further than we
have been able to do so far.

Let us say that in the sample of radioactive material there are N nuclei
which have not decayed at a certain time, t. So what happens in the next
brief period of time? Some nuclei will decay for sure. But how many?

On the basis of our reasoning above we can say that the number which
will decay will depend on overall number of nuclei, N, and also on the
length of the brief period of time. In other words the more nuclei there
are the more will decay and the longer the time period the more nuclei
will decay. Let us denote the number which will have decayed as **dN**
and the small time interval as **dt**.

So we have reasoned that the number of radioactive nuclei which will
decay during the time interval from t to t+dt must be proportional to N
and to dt. In symbols therefore:

```{=html}
<div class="center">
```
$-dN \propto N \cdot dt\,\!$

```{=html}
</div>
```
the minus sign indicating that N is decreasing.

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

```{=html}
<div class="center">
```
$-dN = \lambda N \cdot dt\,\!$

```{=html}
</div>
```
where the constant of proportionality, λ (named lambda) is called the
**Decay Constant**.

Dividing across by N we can rewrite this equation as:

```{=html}
<div class="center">
```
$-\frac{dN}{N} = \lambda \cdot dt$

```{=html}
</div>
```
So this equation describes the situation for any brief time interval,
dt. To find out what happens for all periods of time we simply add up
what happens in each brief time interval. In other words we
integrate the above equation. Expressing this
more formally we can say that for the period of time from t = 0 to any
later time t, the number of radioactive nuclei will decrease from N~0~
to N~t~, so that:

```{=html}
<div class="center">
```
$-\int_{N_0}^{N_t} \frac{dN}{N} = \lambda \int_{0}^t dt$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \ln \left ( \frac{N_t}{N_0} \right ) = -\lambda t$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \frac{N_t}{N_0} = \text{exp}\,(-\lambda t)$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore N_t = N_0 \text{exp}\,(-\lambda t)$

```{=html}
</div>
```
This final expression is known as the **Radioactive Decay Law**. It
tells us that the number of radioactive nuclei will decrease in an
exponential fashion with time with the rate of decrease being controlled
by the Decay Constant.

Before looking at this expression in further detail let us review the
mathematics which we used above. First of all we used integral calculus
to figure out what was happening over a period of time by integrating
what we knew would occur in a brief interval of time. Secondly we used a
calculus relationship that the

```{=html}
<div class="center">
```
$\int \frac{dx}{x} = \ln x$

```{=html}
</div>
```
where ln *x* represents the natural
logarithm
of *x*. And thirdly we used the definition of logarithms that when

```{=html}
<div class="center">
```
$\ln x = y\,\!$

```{=html}
</div>
```
then,

```{=html}
<div class="center">
```
$x = \text{exp}\ y\,\!$

```{=html}
</div>
```
Now, to return to the Radioactive Decay Law. The Law tells us that the
number of radioactive nuclei will decrease with time in an exponential
fashion with the rate of decrease being controlled by the Decay
Constant. The Law is shown in graphical form in the figure below:

![](NM4_11a.gif "NM4_11a.gif")

The graph plots the number of radioactive nuclei at any time, *N*~*t*~,
against time, *t*. We can see that the number of radioactive nuclei
decreases from *N*~0~ that is the number at *t* = 0 in a rapid fashion
initially and then more slowly in the classic exponential manner.

The influence of the Decay Constant can be seen in the following figure:

![](NM4_12a.gif "NM4_12a.gif")

All three curves here are exponential in nature, only the Decay Constant
is different. Notice that when the Decay Constant has a low value the
curve decreases relatively slowly and when the Decay Constant is large
the curve decreases very quickly.

The Decay Constant is characteristic of individual radionuclides. Some
like uranium-238 have a small value and the material therefore decays
quite slowly over a long period of time. Other nuclei such as
technetium-99m have a relatively large Decay Constant and they decay far
more quickly.

It is also possible to consider the Radioactive Decay Law from another
perspective by plotting the logarithm of N~t~ against time. In other
words from our analysis above by plotting the expression:

```{=html}
<div class="center">
```
$\ln \left ( \frac{N_t}{N_0} \right ) = - \lambda t$

```{=html}
</div>
```
in the form

```{=html}
<div class="center">
```
$\ln N_t = -\lambda t + \ln N_0\,\!$

```{=html}
</div>
```
Notice that this expression is simply an equation of the form *y* =
*mx* + *c* where *m* = -l and *c* = ln *N*~0~. As a result it is the
equation of a straight line of slope -l as shown in the following
figure. Such a plot is sometimes useful when we wish to consider a
situation without the complication of the direct exponential behaviour.

![](NM4_15a.gif "NM4_15a.gif")

## Half-Life

Most of us have not been taught to think instinctively in terms of
logarithmic or exponential terms even though many natural phenomena
display exponential behaviours. Most of the forms of thinking which we
have been taught in school are based on linear changes and as a result
it is rather difficult for us to grasp the Radioactive Decay Law
intuitively. For this reason an indicator is usually derived from the
law which helps us think more clearly about what is going on.

This indicator is called the **Half Life** and it expresses the length
of time it takes for the radioactivity of a radioisotope to decrease by
a factor of two. From a graphical point of view we can say that when:

```{=html}
<div class="center">
```
$N_t = \frac{N_0}{2}$

```{=html}
</div>
```
the time taken is the Half Life:

![](NM4_18a.gif "NM4_18a.gif")

Note that the half-life does not express how
long a material will remain radioactive but simply the length of time
for its radioactivity to halve. Examples of the half lives of some
radioisotopes are given in the following table. Notice that some of
these have a relatively short half life. These tend to be the ones used
for medical diagnostic purposes because they do not remain radioactive
for very long following administration to a patient and hence result in
a relatively low radiation dose.

  Radioisotope   Half Life (approx.)
  -------------- ---------------------
  ^81m^Kr        13 seconds
  ^99m^Tc        6 hours
  ^131^I         8 days
  ^51^Cr         1 month
  ^137^Cs        30 years
  ^241^Am        462 years
  ^226^Ra        1620 years
  ^238^U         4.51 x 10^9^ years

But they do present a logistical problem when we wish to use them when
there may not be a radioisotope production facility nearby. For example
suppose we wish to use ^99m^Tc for a patient study and the nearest
nuclear facility for making this isotope is 5,000 km away. The
production facility could be in Sydney and the patient could be in Perth
for instance. After making the isotope at the nuclear plant it would be
decaying with a half life of 6 hours. So we put the material on a truck
and drive it to Sydney airport. The isotope would be decaying as the
truck sits in Sydney traffic then decaying still more as it waits for a
plane to take it to Perth. Then decaying more as it is flown across to
Perth and so on. By the time it gets to our patient it will have
substantially reduced in radioactivity possibly to the point of being
useless for the patient\'s investigation. And what about the problem if
we needed to use ^81m^Kr instead of ^99m^Tc for our patient? You will
see in another chapter of
this book that logistical challenges such as this have given rise to
quite innovative solutions. More about that later!

You can appreciate from the table above that other isotopes have a very
long half lives. For example ^226^Ra has a half life of over 1,500
years. This isotope has been used in the past for therapeutic
applications in medicine. Think about the logistical problems here. They
obviously do not relate to transporting the material from the point of
production to the point of use. But they relate to how the material is
kept following its arrival at the point of use. We must have a storage
facility so that the material can be kept safely for a long period of
time. But for how long? A general rule of thumb for the quantities of
radioactivity used in medicine is that the radioactivity will remain
significant for about 10 half lives. So we would have to have a safe
environment for storage of the ^226^Ra for about 16,000 years! This
storage facility would have to be secure from many unforeseeable events
such as earthquakes, bombing etc. and be kept in a manner which our
children\'s, children\'s children can understand. A very serious
undertaking indeed!

## Relationship between the Decay Constant and the Half Life

On the basis of the above you should be able to appreciate that there is
a relationship between the Decay Constant and the Half Life. For example
when the Decay Constant is small the Half Life should be long and
correspondingly when the Decay Constant is large the Half Life should be
short. But what exactly is the nature of this relationship?

We can easily answer this question by using the definition of Half Life
and applying it to the Radioactive Decay Law.

Once again the law tells us that at any time, *t*:

```{=html}
<div class="center">
```
$N_t = N_0\ \text{exp}\,(-\lambda t)\,\!$

```{=html}
</div>
```
and the definition of Half Life tells us that:

```{=html}
<div class="center">
```
$N_t = \frac{N_0}{2}$

```{=html}
</div>
```
when

```{=html}
<div class="center">
```
$t = t_{\frac{1}{2}}$

```{=html}
</div>
```
We can therefore re-write the Radioactive Decay Law by substituting for
*N*~t~ and *t* as follows:

```{=html}
<div class="center">
```
$\frac{N_0}{2} = N_0\ \text{exp}\,(-\lambda t_{\frac{1}{2}})$

```{=html}
</div>
```
Therefore

```{=html}
<div class="center">
```
$\frac{1}{2} = \text{exp}\,(-\lambda t_{\frac{1}{2}})$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore 2^{-1} = \text{exp}\,(-\lambda t_{\frac{1}{2}})$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \ln 2^{-1} = -\lambda t_{\frac{1}{2}}$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \ln 2 = \lambda t_{\frac{1}{2}}$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore 0.693 = \lambda t_{\frac{1}{2}}$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$t_{\frac{1}{2}} = \frac{0.693}{\lambda}$

```{=html}
</div>
```
and

```{=html}
<div class="center">
```
$\lambda = \frac{0.693}{t_{\frac{1}{2}}}$

```{=html}
</div>
```
These last two equations express the relationship between the Decay
Constant and the Half Life. They are very useful as you will see when
solving numerical questions relating to radioactivity and usually form
the first step in solving a numerical problem.

## Units of Radioactivity

The SI or metric unit of radioactivity is named after Henri Becquerel,
in honour of his discovery of radioactivity, and is called the
**becquerel** with the symbol Bq. The
becquerel is defined as the quantity of
radioactive substance that gives rise to a decay rate of 1 decay per
second.

In medical diagnostic work 1 Bq is a rather small amount of
radioactivity. Indeed it is easy to remember its definition if you think
of it as a *buggerall* amount of radioactivity. For this reason the
kilobecquerel (kBq) and megabecquerel (MBq) are more frequently used.

The traditional unit of radioactivity is named after Marie Curie and is
called the **curie**, with the symbol Ci. The
curie is defined as the amount of radioactive
substance which gives rise to a decay rate of 3.7 x 10^10^ decays per
second. In other words 37 thousand, million decays per second which as
you might appreciate is a substantial amount of radioactivity. For
medical diagnostic work the millicurie (mCi) and the microcurie (µCi)
are therefore more frequently used.

Why two units? It in essence like all other units of measurement depends
on what part of the world you are in. For example the kilometer is
widely used in Europe and Australia as a unit of distance and the mile
is used in the USA. So if you are reading an American textbook you are
likely to find the curie used as the unit of radioactivity, if you are
reading an Australian book it will most likely refer to becquerels and
both units might be used if you are reading a European book. You will
therefore find it necessary to know and understand both units.

## Multiple Choice Questions

Click
`<b>`{=html}here`</b>`{=html}
to access an MCQ on the Radioactive Decay Law.

## Questions

Three 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 Radioactive
Decay Law as well as your understanding of the concept of Half Life. The
second question is a lot more challenging and will help you relate the
Radioactive Decay Law to the number of radioactive nuclei which are
decaying in a sample of radioactive material. The third question will
help you understand the approach used in the second question by asking a
similar question from a slightly different perspective.

**Question 1**

\(a\) The half-life of ^99m^Tc is 6 hours. After how much time will
1/16th of the radioisotope remain?

\(b\) Verify your answer by another means.

**Answer**:

:   \(a\) Starting with the relationship we established earlier between
    the Decay Constant and the Half Life we can calculate the Decay
    Constant as follows:

```{=html}
<div class="center">
```
$\lambda = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{6} = 0.1155\ \text{hr}^{-1}$

```{=html}
</div>
```

:   Now applying the Radioactive Decay Law,

```{=html}
<div class="center">
```
$N_t = N_0\ \text{exp}\,(-\lambda t)\,\!$

```{=html}
</div>
```

:   we can re-write it in the form:

```{=html}
<div class="center">
```
$\frac{N_t}{N_0} = \text{exp}\,(-\lambda t)$

```{=html}
</div>
```

:   The question tells us that N~0~ has reduced to 1/16th of its value,
    that is:

```{=html}
<div class="center">
```
$\frac{N_t}{N_0} = \frac{1}{16}$

```{=html}
</div>
```

:   Therefore

```{=html}
<div class="center">
```
$\frac{1}{16} = \text{exp}\,(-0.1155t)$

```{=html}
</div>
```

:   which we need to solve for t. One way of doing this is as follows:

```{=html}
<div class="center">
```
$16^{-1} = \text{exp}\,(-0.1155t)$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore -\ln 16 = -0.1155t\,\!$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$t = \frac{\ln 16}{0.1155} = 24\ \text{hours}$

```{=html}
</div>
```

:   So it will take 24 hours until 1/16th of the radioactivity remains.

```{=html}
<!-- -->
```

:   \(b\) A way in which this answer can be verified is by using the
    definition of Half Life. We are told that the Half Life of ^99m^Tc
    is 6 hours. Therefore after six hours half of the radioactivity
    remains.

```{=html}
<!-- -->
```

:   Therefore after 12 hours a quarter remains; after 18 hours an eighth
    remains and after 24 hours one sixteenth remains. And we arrive at
    the same answer as in part (a). So we must be right!

```{=html}
<!-- -->
```

:   Note that this second approach is useful if we are dealing with
    relatively simple situations where the radioactivity is halved,
    quartered and so on. But supposing the question asked how long would
    it take for the radioactivity to decrease to a tenth of its initial
    value. Deduction from the definition of half life is rather more
    difficult in this case and the mathematical approach used for
    part (a) above will yield the answer more readily.

**Question 2**

Find the radioactivity of a 1 g sample of ^226^Ra given that *t*~1/2~:
1620 years and Avogadro\'s Number: 6.023 x 10^23^.

**Answer**:

:   We can start the answer like we did with Question 1(a) by
    calculating the Decay Constant from the Half Life using the
    following equation:

```{=html}
<div class="center">
```
$\lambda = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{1620} = 4.28 \cdot 10^{-4}\ \text{year}^{-1}$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \lambda = 1.36 \cdot 10^{-11} s^{-1}$

```{=html}
</div>
```

:   Note that the length of a year used in converting from \'per year\'
    to \'per second\' above is 365.25 days to account for leap years. In
    addition the reason for converting to units of \'per second\' is
    because the unit of radioactivity is expressed as the number of
    nuclei decaying per second.

```{=html}
<!-- -->
```

:   Secondly we can calculate that 1 g of ^226^Ra contains:

```{=html}
<div class="center">
```
$N = \frac{(\text{Avogadro No.})(\text{Mass})}{\text{Mass Number}} = \frac{(6.023 \cdot 10^{23})(1g)}{226} = 2.7 \cdot 10^{21}\ \text{nuclei}$

```{=html}
</div>
```

:   Thirdly we need to express the Radioactive Decay Law in terms of the
    number of nuclei decaying per unit time. We can do this by
    differentiating the equation as follows:

```{=html}
<div class="center">
```
$N = N_0\ \text{exp}\,(-\lambda t)$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \frac{dN}{dt} = N_0 \cdot -\lambda\ \text{exp}\,(-\lambda t) = -\lambda N_0\ \text{exp}\,(-\lambda t)$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \frac{dN}{dt} = -\lambda N$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \left | \frac{dN}{dt} \right | = \lambda N$

```{=html}
</div>
```

:   The reason for expressing the result above in absolute terms is to
    remove the minus sign in that we already know that the number is
    decreasing.

```{=html}
<!-- -->
```

:   We can now enter the data we derived above for λ and *N*:

```{=html}
<div class="center">
```
$\left | \frac{dN}{dt} \right | = (1.36 \cdot 10^{-11})(2.7 \cdot 10^{21})$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \left | \frac{dN}{dt} \right | = 3.6 \cdot 10^{10}\ \text{decays per second}$

```{=html}
</div>
```

:   So the radioactivity of our 1 g sample of radium-226 is
    approximately 1 Ci.

```{=html}
<!-- -->
```

:   This is not a surprising answer since the definition of the
    **curie** was originally conceived as the radioactivity of 1 g of
    radium-226!

**Question 3**

What is the minimum mass of ^99m^Tc that can have a radioactivity of 1
MBq? Assume the half-life is 6 hours and that Avogadro\'s Number is
6.023 x 10^23^.

**Answer**

:   Starting again with the relationship between the Decay Constant and
    the Half Life:

```{=html}
<div class="center">
```
$\lambda = \frac{0.693}{6} = 0.1155\ \text{hour}^{-1} = 3.21 \cdot 10^{-5}$

```{=html}
</div>
```

:   Secondly the question tells us that the radioactivity is 1 MBq.
    Therefore since 1 MBq = 1 x 10^6^ decays per second,

```{=html}
<div class="center">
```
$\left | \frac{dN}{dt} \right | = \lambda N = 1 \cdot 10^6\ \text{dps}$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore N = \frac{\left | \frac{dN}{dt} \right |}{\lambda} = \frac{1 \cdot 10^6}{3.21 \cdot 10^{-5}} = 3.116 \cdot 10^{10}$

```{=html}
</div>
```

:   Finally the mass of these nuclei can be calculated as follows:

```{=html}
<div class="center">
```
$\text{Mass of N nuclei} = \frac{(\text{No. of Nuclei})(\text{Mass No.})}{\text{Avogadro Number}}$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$= \frac{(3.116 \cdot 10^{10})(99)}{6.023 \cdot 10^{23}} = 5.122 \cdot 10^{-12}\ \text{g}$

```{=html}
</div>
```

:   In other words a mass of just over five picograms of ^99m^Tc can
    emit one million gamma-rays per second. The result reinforces an
    important point that you will learn about radiation protection which
    is that you should treat radioactive materials just like you would
    handle pathogenic bacteria!


# Basic Physics of Nuclear Medicine/Units of Radiation Measurement

!`<span style="color:white;">`{=html}.`</span>`{=html}{width="256"}

## Introduction

This is the fourth chapter of a wikibook entitled Basic Physics of
Nuclear Medicine.

After that rather long and detailed chapter we have just finished we
will now proceed at a more leisurely pace for a short treatment of some
of the more common units of measurement used in this field.

Before we do so however it is useful to consider the typical radiation
environment. By doing so we will gain an appreciation of the various
quantities that can be measured before considering the units which are
used to express such measurements. So, we will first of all consider a
typical radiation situation and then go on to consider the various units
of measurement.

## A Typical Radiation Situation

A typical radiation set-up is shown in the figure below. Firstly there
is a **source** of radiation, secondly a radiation **beam** and thirdly
some **material** which absorbs the radiation. So the quantities which
can be measured are associated with the source, the radiation beam and
the absorber.

```{=html}
<div class="center">
```
![](NM5_1.gif "NM5_1.gif")

```{=html}
</div>
```
This type of environment could be one where the radiation from the
source is used to irradiate a patient (that is the absorber) for
diagnostic purposes where we would place a device behind the patient for
producing an image or for therapeutic purposes where the radiation is
intended to cause damage to a specific region of a patient. It is also a
situation where we as an absorber may be working with a source of
radiation.

## The Radiation Source

When the radiation source is a radioactive one the quantity that is
typically measured is the radioactivity of the source. We saw in the
previous chapter that the units used to express radioactivity are the
**becquerel** (SI unit) and the **curie** (traditional unit).

## The Radiation Beam

The characteristic of a radiation beam that is typically measured is
called the **Radiation Exposure**. This quantity expresses how much
ionisation the beam causes in the air through which it travels.

We will see in the following
chapter
that one of the major things that happens when radiation encounters
matter is that ions are formed -- air being the form
of matter it encounters in this case. So the radiation exposure produced
by a radiation beam is expressed in terms of the amount of ionisation
which occurs in air.

A straight-forward way of measuring such ionisation is to determine the
amount of electric charge which is produced. You will remember from your
high school physics that the SI unit of electric charge is the
**coulomb**.

The SI unit of radiation exposure is the **coulomb per kilogram** -- and
is given the symbol C kg^-1^. It is defined as the quantity of X- or
gamma-rays such that the associated electrons emitted per kilogram of
air at standard temperature and pressure (STP) produce ions carrying 1
coulomb of electric charge.

The traditional unit of radiation exposure is the **roentgen**, named in
honour of Wilhelm Roentgen (who
discovered X-rays) and is given the symbol R. The
roentgen is defined as the quantity of X- or
gamma-rays such that the associated electrons emitted per kilogram of
air at STP produce ions carrying 2.58 x 10^-4^ coulombs of electric
charge.

So 1 R is a small exposure relative to 1 C kg^-1^ -- in fact it is 3,876
times smaller.

Note that this unit is confined to radiation beams consisting of X-rays
or gamma-rays.

Often it is not simply the exposure that is of interest but the exposure
rate, that is the exposure per unit time. The units which tend to be
used in this case are the C kg^-1^ s^-1^ and the R hr^-1^.

## The Absorber

Energy is deposited in the absorber when radiation interacts with it. It
is usually quite a small amount of energy but energy nonetheless. The
quantity that is measured is called the **Absorbed Dose** and it is of
relevance to all types of radiation be they X- or gamma-rays, alpha- or
beta-particles.

The SI unit of absorbed dose is called the **gray**, named after a
famous radiobiologist, LH Gray, and is
given the symbol Gy. The gray "wikilink") is defined as
the absorption of 1 joule of radiation energy per kilogram of material.
So when 1 joule of radiation energy is absorbed by a kilogram of the
absorber material we say that the absorbed dose is 1 Gy.

The traditional unit of absorbed dose is called the **rad**, which
supposedly stands for Radiation Absorbed
Dose "wikilink"). It is defined as the absorption of
10^-2^ joules of radiation energy per kilogram of material.

As you can figure out 1 Gy is equal to 100 rad.

There are other quantities derived from the gray and the rad which
express the biological effects of such absorbed radiation energy when
the absorber is living matter -- human tissue for example. These
quantities include the Equivalent Dose,
`<span style="color:red;">`{=html}H`</span>`{=html}, and the Effective
Dose,
`<span style="color:red;">`{=html}E`</span>`{=html}. The Equivalent Dose
is based on estimates of the ionization capability of the different
types of radiation which are called **Radiation Weighting Factors**,
`<span style="color:red;">`{=html}w~R~`</span>`{=html}, such that

```{=html}
<div class="center">
```
`<span style="color:red;">`{=html}`<big>`{=html}H = w~R~
D`</big>`{=html}`</span>`{=html}

```{=html}
</div>
```
where `<span style="color:red;">`{=html}D`</span>`{=html} is the
absorbed dose. The Effective Dose includes
`<span style="color:red;">`{=html}w~R~`</span>`{=html} as well as
estimates of the sensitivity of different tissues called **Tissue
Weighting Factors**,
`<span style="color:red;">`{=html}w~T~`</span>`{=html}, such that

```{=html}
<div class="center">
```
`<span style="color:red;">`{=html}`<big>`{=html}E = Σ w~T~
H`</big>`{=html}`</span>`{=html}

```{=html}
</div>
```
where the summation,
`<span style="color:red;">`{=html}Σ`</span>`{=html}, is over all the
tissue types involved. Both the Equivalent Dose and the Effective Dose
are measured in derived SI units called sieverts
(Sv).

Let us pause here for a bit to ponder on the use of the term *dose*. It
usually has a medical connotation in that we can say that someone had a
dose of the \'flu, or that the doctor prescribed a certain dose of a
drug. What has it to do with the deposition of energy by a beam of
radiation in an absorber? It could have something to do with the initial
applications of radiation in the early part of the 20^th^ century when
it was used to treat numerous diseases. As a result we can speculate
that the term has stayed in the vernacular of the field. It would be
much easier to use a term like absorbed radiation energy since we are
talking about the deposition of energy in an absorber. But this might
make the subject just a little too simple!

## Specific Gamma Ray Constant

A final quantity is worth mentioning with regard to radiation units.
This is the Specific Gamma-Ray Constant for a radioisotope. This
quantity is an amalgam of the quantities we have already covered and
expresses the exposure rate produced by the gamma-rays emitted from a
radioisotope.

It is quite a useful quantity from a practical viewpoint when we are
dealing with a radioactive source which emits gamma-rays. Supposing you
are using a gamma-emitting radioactive source (for example ^99m^Tc or
^137^Cs) and you will be standing at a certain distance from this source
while you are working. You most likely will be interested in the
exposure rate produced by the source from a radiation safety point of
view. This is where the Specific Gamma-Ray Constant comes in.

It is defined as the exposure rate per unit activity at a certain
distance from a source. The SI unit is therefore the

```{=html}
<div class="center">
```
`<span style="color:red;">`{=html}`<big>`{=html}C kg^-1^ s^-1^ Bq^-1^ at
1 m `</big>`{=html}`</span>`{=html},

```{=html}
</div>
```
and the traditional unit is the

```{=html}
<div class="center">
```
`<span style="color:red;">`{=html}`<big>`{=html}R hr^-1^ mCi^-1^ at 1 cm
`</big>`{=html}`</span>`{=html} .

```{=html}
</div>
```
These units of measurement are quite cumbersome and a bit of a mouthful.
It might have been better if they were named after some famous scientist
so that we could call the SI unit 1 **smith** and the traditional unit 1
**jones** for example. But again things are not that simple!

## The Inverse Square Law

Before we finish this chapter we are going to consider what happens as
we move our absorber away from the radiation source. In other words we
are going to think about the influence of distance on the intensity of
the radiation beam. You will find that a useful result emerges from this
that has a very important impact on radiation safety.

The radiation produced in a radioactive source is emitted in all
directions. We can consider that spheres of equal radiation intensity
exist around the source with the number of photons/particles spreading
out as we move away from the source.

Consider an area on the surface of one of these spheres and assume that
there are a certain number of photons/particles passing though it. If we
now consider a sphere at a greater distance from the source the same
number of photons/particles will now be spread out over a bigger area.
Following this line of thought it is easy to appreciate that the
radiation intensity, I will decrease with the square of the distance, r
from the source, i.e.

```{=html}
<div class="center">
```
$I \propto \frac{1}{r^2}$

```{=html}
</div>
```
This effect is known as the Inverse Square
Law. As a result if we double the
distance from a source, we reduce the intensity by a factor of two
squared, that is 4. If we triple the distance the intensity is reduced
by a factor of 9, that is three squared, and so on.

This is a very useful piece of information if you are working with a
source of radiation and are interested in minimising the dose of
radiation you will receive.

## External links

-   Radiation and
    Risk -- covers
    the effect of radiation, how risks are determined, comparison of
    radiation with other risks and radiation doses.
-   Radiation Effects
    Overview -- results
    of studies of victims of nuclear bombs including early effects on
    survivors, effects on the in utero exposed, and late effects on the
    survivors -- from the Radiation Effects Research Foundation, a
    cooperative Japan-United States Research Organization.
-   The Radiation and Health Physics Home
    Page -- all you ever wanted to know
    about radiation but were afraid to ask\....with hundreds of WWW
    links -- from the Student Chapter of the Health Physics Society,
    University of Michigan containing sections on general information,
    regulatory Information, professional organizations and societies,
    radiation specialties, health physics research and education.
-   What You Need to Know about
    Radiation
    -- to protect yourself to protect your family to make reasonable
    social and political choices -- covers sources of radiation and
    radiation protection -- by Lauriston S. Taylor.


# Basic Physics of Nuclear Medicine/Interaction of Radiation with Matter

!`<span style="color:white;">`{=html}.`</span>`{=html}{width="150"}

We have focussed in previous chapters on the source of radiation and the
types of radiation. We are now in a position to consider what happens
when this radiation interacts with matter. Our main reason for doing
this is to find out what happens to the radiation as it passes through
matter and also to set ourselves up for considering how it interacts
with living tissue and how to detect radiation. Since all radiation
detectors are made from some form of matter it is useful to first of all
know how radiation interacts so that we can exploit the effects in the
design of such detectors in subsequent chapters of this wikibook.

Before we do this let us first remind ourselves of the physical
characteristics of the major types of radiation. We have covered this
information in some detail
earlier
and it is summarised in the table below for convenience.

We will now consider the passage of each type of radiation through
matter with most attention given to gamma-rays because they are the most
common type used in nuclear medicine. One of the main effects that you
will notice irrespective of the type of radiation is that ions are
produced when radiation interacts with matter. It is for this reason
that it is called ionizing radiation.

+----------------+----------------+----------------+----------------+
| Radiation      | Mass           | Electric       | Velocity       |
|                |                | Charge         |                |
+================+================+================+================+
| Alpha          | ```{=html}     | ```{=html}     | ```{=html}     |
| Particles      | <div c         | <div c         | <div c         |
|                | lass="center"> | lass="center"> | lass="center"> |
|                | ```            | ```            | ```            |
|                | relatively     | double         | relatively     |
|                | heavy          | positive       | slow           |
|                |                |                |                |
|                | ```{=html}     | ```{=html}     | ```{=html}     |
|                | </div>         | </div>         | </div>         |
|                | ```            | ```            | ```            |
+----------------+----------------+----------------+----------------+
| Beta Particles | ```{=html}     | ```{=html}     | ```{=html}     |
|                | <div c         | <div c         | <div c         |
|                | lass="center"> | lass="center"> | lass="center"> |
|                | ```            | ```            | ```            |
|                | about 8,000    | negative       | less than the  |
|                | times lighter  |                | velocity of    |
|                |                | ```{=html}     | light          |
|                | ```{=html}     |
         |                |
|                |
         | ```            | ```{=html}     |
|                | ```            |                | </div>         |
|                |                |                | ```            |
+----------------+----------------+----------------+----------------+
| Gamma Rays     | ```{=html}     | ```{=html}     | ```{=html}     |
|                | <div c         | <div c         | <div c         |
|                | lass="center"> | lass="center"> | lass="center"> |
|                | ```            | ```            | ```            |
|                | None           | None           | 3 x 10^8^ m/s  |
|                |                |                | in free space  |
|                | ```{=html}     | ```{=html}     |                |
|                |
         |
         | ```{=html}     |
|                | ```            | ```            | </div>         |
|                |                |                | ```            |
+----------------+----------------+----------------+----------------+

Before we start though you might find an analogy useful to help you with
your thinking. This analogy works on the basis of thinking about matter
as an enormous mass of atoms (that is nuclei with orbiting electrons)
and that the radiation is a particle/photon passing through this type of
environment. So the analogy to think about is a spaceship passing
through a meteor storm like you might see in a science-fiction movie
where the spaceship represents the radiation and the meteors represent
the atoms of the material through which the radiation is passing. One
added feature to bring on board however is that our spaceship sometimes
has an electric charge depending on the type of radiation it represents.

## Alpha Particles

We can see from the table above that
alpha-particles have a double positive
charge and we can therefore easily appreciate that they will exert
considerable electrostatic attraction on the outer orbital electrons of
atoms near which they pass. The result is that some electrons will be
attracted away from their parent atoms and that ions will be produced.
In other words ionizations occur.

We can also appreciate from the table that alpha-particles are quite
massive relative to the other types of radiation and also to the
electrons of atoms of the material through which they are passing. As a
result they travel in straight lines through matter except for rare
direct collisions with nuclei of atoms along their path.

A third feature of relevance here is the energy with which they are
emitted. This energy in the case of alpha-particles is always distinct.
For example ^221^Ra emits an alpha-particle with an energy of 6.71 MeV.
Every alpha-particle emitted from this radionuclide has this energy.
Another example is ^230^U which emits three alpha-particles with
energies of 5.66, 5.82, 5.89 MeV.

Finally it is useful to note that alpha-particles are very damaging
biologically and this is one reason why they are not used for in-vivo
diagnostic studies. We will therefore not be considering them in any
great detail in this wikibook.

## Beta Particles

We can see from the table that
beta-particles have a negative electric
charge. Notice that positrons are not considered here since as we noted
in chapter
2 these
particles do not last for very long in matter before they are
annihilated. Beta-minus particles last considerably longer and are
therefore the focus of our attention here.

Because of their negative charge they are attracted by nuclei and
repelled by electron clouds as they pass through matter. The result once
again without going into great detail is ionization.

The path of beta-particles in matter is often described as being
tortuous, since they tend to ricochet from atom to atom.

A final and important point to note is that the energy of beta-particles
is never found to be distinct in contrast to the alpha-particles above.
The energies of the beta-particles from a radioactive source forms a
spectrum up to a maximum energy -- see figure below. Notice from the
figure that a range of energies is present and features such as the mean
energy, E~mean~, or the maximum energy, E~max~, are quoted.

```{=html}
<div class="center">
```
![](NM6_1.gif "NM6_1.gif"){width="280"}

```{=html}
</div>
```
The question we will consider here is: why should a spectrum of energies
be seen? Surely if a beta-particle is produced inside a nucleus when a
neutron is converted into a proton, a single distinct energy should
result. The answer lies in the fact that two particles are actually
produced in beta-decay. We did not cover this in our treatment in
chapter
2 for
fear of complicating things too much at that stage of this wikibook. But
we will cover it here briefly for the sake of completeness.

The second particle produced in beta-decay is called a
neutrino and was named by Enrico
Fermi. It is quite a mysterious particle
possessing virtually no mass and carrying no charge, though we are still
researching its properties today. The difficulty with them is that they
are very hard to detect and this has greatly limited our knowledge about
them so far.

The beta-particle energy spectrum can be explained by considering that
the energy produced when a neutron is converted to a proton is shared
between the beta-particle and the anti-neutrino. Sometimes all the
energy is given to the beta-particle and it receives the maximum energy,
E~max~. But more often the energy is shared between them so that for
example the beta-particle has the mean energy, E~mean~ and the neutrino
has the remainder of the energy.

Finally it is useful to note that beta-particles are quite damaging
biologically and this is one reason why they are not used for in-vivo
diagnostic studies. We will therefore not consider them in any great
detail in this wikibook.

## Gamma Rays

Since we have been talking about energies above, let us first note that
the energies of gamma-rays emitted from a
radioactive source are always distinct. For example ^99m^Tc emits
gamma-rays which all have an energy of 140 keV and ^51^Cr emits
gamma-rays which have an energy of 320 keV.

Gamma-rays have many modes of interaction with matter. Those which have
little or no relevance to nuclear medicine imaging are:

::\* Mössbauer Effect

::\* Coherent Scattering

::\* Pair Production

and will not be described here.

Those which are very important to nuclear medicine imaging, are the
Photoelectric Effect and the
Compton Effect. We will consider each
of these in turn below. Note that the effects described here are also of
relevance to the interaction of X-rays with matter since as we have
noted before X-rays and gamma-rays are essentially the same entities. So
the treatment below is also of relevance to radiography.

#### Photoelectric Effect

:   When a gamma-ray collides with an orbital electron of an atom of the
    material through which it is passing it can transfer all its energy
    to the electron and cease to exist -- see figure below. On the basis
    of the Principle of Conservation of
    Energy we can deduce that the
    electron will leave the atom with a kinetic energy equal to the
    energy of the gamma-ray less that of the orbital binding energy.
    This electron is called a **photoelectron**.

```{=html}
<div class="center">
```
![](NM6_2.gif "NM6_2.gif"){width="320"}

```{=html}
</div>
```

:   Note that an ion results when the photoelectron leaves the atom.
    Also note that the gamma-ray energy is totally absorbed in the
    process.

```{=html}
<!-- -->
```

:   Two subsequent points should also be noted. Firstly the
    photoelectron can cause ionisations along its track in a similar
    manner to a beta-particle. Secondly X-ray
    emission can occur when the
    vacancy left by the photoelectron is filled by an electron from an
    outer shell of the atom. Remember that we came across this type of
    feature before when we dealt with Electron Capture in chapter
    2.

#### Compton Effect

:   This type of effect is somewhat akin to a cue ball hitting a
    coloured ball on a pool table. Here a gamma-ray transfers only part
    of its energy to a valance electron which is essentially free -- see
    figure below. Notice that the electron leaves the atom and may act
    like a beta-particle and that the gamma-ray deflects off in a
    different direction to that with which it approached the atom. This
    deflected or scattered gamma-ray can undergo further Compton Effects
    within the material.

```{=html}
<!-- -->
```

:   Note that this effect is sometimes called **Compton Scattering**.

```{=html}
<div class="center">
```
![](NM6_3.gif "NM6_3.gif"){width="320"}

```{=html}
</div>
```
The two effects we have just described give rise to both absorption and
scattering of the radiation beam. The overall effect is referred to as
**attenuation** of gamma-rays. We will investigate this feature from an
analytical perspective in the following chapter. Before we do so, we\'ll
briefly consider the interaction of radiation with living matter.

## Radiation Biology

It is well known that exposure to ionizing radiation can result in
damage to living tissue. We\'ve already described the initial atomic
interactions. What\'s important in radiation biology is that these
interactions may trigger complex chains of biomolecular events and
consequent biological damage.

We\'ve seen above that the primary means by which ionizing radiations
lose their energy in matter is by ejection of orbital electrons. The
loss of orbital electrons from the atom leaves it positively charged.
Other interaction processes lead to **excitation** of the atom rather
than ionization. Here, an outer valence electron receives sufficient
energy to overcome the binding energy of its shell and moves further
away from the nucleus to an orbit that is not normally occupied. This
type of effect alters the chemical force that binds atoms into molecules
and a regrouping of the affected atoms into different molecular
structures can result. That is, excitation is an indirect method of
inducing chemical change through the modification of individual atomic
bonds.

Ionizations and excitations can give rise to unstable chemical species
called free radicals. These are atoms and
molecules in which there are unpaired electrons. They are chemically
very reactive and seek stability by bonding with other atoms and
molecules. Changes to nearby molecules can arise because of their
production.

But, let\'s go back to the interactions themselves for the moment\.....

In the case of X- and gamma-ray interactions, the energy of the photons
is usually transferred by collisions with orbital electrons, e.g. via
photoelectric and Compton effects. These radiations are capable of
penetrating deeply into tissue since their interactions depend on chance
collisions with electrons. Indeed, nuclear medicine imaging is only
possible when the energy of the gamma-rays is sufficient for complete
emission from the body, but low enough to be detected.

The interaction of charged particles (e.g. alpha and beta particles), on
the other hand, can be by collisions with atomic electrons and also via
attractive and repulsive electrostatic forces. The rate at which energy
is lost along the track of a charged particle depends therefore on the
square of the charge on that particle.
That is, the greater the particle charge, the greater the probability of
it generating ion pairs along its track. In addition, a longer period of
time is available for electrostatic forces to act when a charged
particle is moving slowly and the ionization probability is therefore
increased as a result.

The situation is illustrated in the following figure where tracks of
charged particles in water are depicted. Notice that the track of the
relatively massive α-particle is a straight line, as we\'ve discussed
earlier in this
chapter,
with a large number of interactions (indicated by the asterisks) per
unit length. Notice also that the tracks for electrons are tortuous, as
we\'ve also discussed
earlier,
and that the number of interactions per unit length is considerably
less.

!Ionizations and excitations along particle tracks in water, for a 5.4
MeV α-particle (top left), for electrons generated following the
absorption of a 1.5 keV X-ray photon (top right) and electrons generated
during the decay of
iodine-125., for electrons generated following the absorption of a 1.5 keV X-ray photon (top right) and electrons generated during the decay of iodine-125."){width="320"}

The Linear Energy Transfer
(**LET**) is defined as the energy released per unit length of the track
of an ionizing particle. A slowly moving, highly charged particle
therefore has a substantially higher LET than a fast, singly charged
particle. An alpha particle of 5 MeV energy and an electron of 1 MeV
energy have LETs, for instance, of 95 and 0.25 keV/μm, respectively. The
ionization density and hence the energy deposition pattern associated
with the heavier charged particle is very much greater than that arising
from electrons, as illustrated in the figure above.

The energy transferred along the track of a charged particle will vary
because the velocity of the particle is likely to be continuously
decreasing. Each interaction removes a small amount of energy from the
particle so that the LET gradually increases along a particle track with
a dramatic increase (called the Bragg Peak)
occurring just before the particle comes to rest.

The International Commission on Radiation Units and
Measurements (**ICRU**) suggest that **lineal
energy** is a better indicator of relative biological effectiveness
(**RBE**). Although lineal energy has the same units as LET (e.g.
keV/μm), it is defined as the:

```{=html}
<div class="center">
```
`<span style="color:green;">`{=html}ratio of the energy deposited in a
volume of tissue to the average diameter of that volume.`</span>`{=html}

```{=html}
</div>
```
Since the microscopic deposition of energy may be quite anisotropic,
lineal energy should be a more appropriate measure of potential damage
than that of LET. The ICRU and the ICRP have
accordingly recommended that the radiation effectiveness of a particular
radiation type should be based on lineal energy in a 1 μm diameter
sphere of tissue. The lineal energy can be calculated for any given
radiation type and energy and a Radiation Weighting
Factor,
(w~R~) can then be determined based on the integrated values of lineal
energy along the radiation track.

All living things on this planet have been exposed to ionizing radiation
since the dawn of time. The current situation for humans is summarized
in the following table:

+----------------------+----------------------+----------------------+
| Source               | Effective Dose       | Comment              |
|                      | (mSv/year)           |                      |
+======================+======================+======================+
| Cosmic radiation     | ```{=html}           | ```{=html}           |
|                      |
 |
 |
|                      | ```                  | ```                  |
|                      | \~0.4                | About 100,000 cosmic |
|                      |                      | ray neutrons and     |
|                      | ```{=html}           | 400,000 secondary    |
|                      | </div>               | cosmic rays          |
|                      | ```                  | penetrate our bodies |
|                      |                      | every hour -- and it |
|                      |                      | increases with       |
|                      |                      | altitude!            |
|                      |                      |                      |
|                      |                      | ```{=html}           |
|                      |                      | </div>               |
|                      |                      | ```                  |
+----------------------+----------------------+----------------------+
| Terrestrial          | ```{=html}           | ```{=html}           |
| radiation            |
 |
 |
|                      | ```                  | ```                  |
|                      | \~0.5                | Over 200 million     |
|                      |                      | gamma-rays pass      |
|                      | ```{=html}           | through our body     |
|                      | </div>               | every hour from      |
|                      | ```                  | sources such as soil |
|                      |                      | and building         |
|                      |                      | materials            |
|                      |                      |                      |
|                      |                      | ```{=html}           |
|                      |                      | </div>               |
|                      |                      | ```                  |
+----------------------+----------------------+----------------------+
| Internal radiation   | ```{=html}           | ```{=html}           |
|                      |
 |
 |
|                      | ```                  | ```                  |
|                      | \~0.3                | About 15 million     |
|                      |                      | ^40^K atoms and      |
|                      | ```{=html}           | about 7,000 natural  |
|                      | </div>               | uranium atoms        |
|                      | ```                  | disintegrate inside  |
|                      |                      | our bodies every     |
|                      |                      | hour, primarily from |
|                      |                      | our diet             |
|                      |                      |                      |
|                      |                      | ```{=html}           |
|                      |                      | </div>               |
|                      |                      | ```                  |
+----------------------+----------------------+----------------------+
| Radon and other      | ```{=html}           | ```{=html}           |
| gases                |
 |
 |
|                      | ```                  | ```                  |
|                      | \~1.3                | About 30,000 atoms   |
|                      |                      | disintegrate inside  |
|                      | ```{=html}           | our lungs every hour |
|                      | </div>               | as a result of       |
|                      | ```                  | breathing            |
|                      |                      |                      |
|                      |                      | ```{=html}           |
|                      |                      | </div>               |
|                      |                      | ```                  |
+----------------------+----------------------+----------------------+

: Adapted with permission from Applied Imaging
Technology.

\
The sum total of this *Natural* Background
Radiation is about 2.5 mSv per year,
with large variations depending on altitude and dietary intake as well
as geological and geographical location.

Its generally considered that repair mechanisms exist in living matter
and that these can be invoked following radiation damage at the
biomolecular level. These mechanisms are likely to have an
evolutionary basis arising as a response to
radiation fluxes generated by natural background sources over the aeons.
Its also known that quite considerable damage to tissues can arise at
quite higher radiation fluxes, even at medical
exposures. Cell death and
transformations to malignant states can result leading to latent periods
of many years before clinical signs of cancer or leukemia, for instance,
become manifest. Further treatment of this vast field of radiation
biology however is beyond our scope
here.

## Practical Radiation Safety

Radiation hazards arise since nuclear medicine involves the handling of
radioactive materials. Although this risk may be small, it remains
important to keep occupational exposures as low as reasonably
achievable. Essential practices for achieving this aim include:

`<b>`{=html}Administration`</b>`{=html}

::\*Maintaining a comprehensive record of all radioactive source
purchases, usage, movement and storage.

::\*Ensuring that any Codes of Safe Practice are adhered to and develop
sensible written protocols and working rules for handling radioisotopes.

::\*Protocols for dealing with minor contamination incidents of the
environment or of staff members must be established. Remember that no
matter how good work practices are, minor accidents or incidents
involving spillage of radioisotopes can take place.
`<b>`{=html}Facility`</b>`{=html}

::\*Storage of radioactive sources in a secure shielded environment.
Specially dedicated facilities are required for the storage, safe
handling, manipulation and dispensing of unsealed radioactive sources.
Storage areas should be designed for both bulk radioisotope and
radioactive waste. Furthermore, radioactive patients should be regarded
as unsealed sources.

::\*Adequate ventilation of any work area. This is particularly
important to minimize the inhalation of Technigas and potentially
volatile radioisotopes such as I-125 and I-131. It is preferable to use
fume hoods when working with volatile materials.

::\*Benches should be manufactured with smooth, hard impervious surfaces
with appropriate splash-backs to allow ready decontamination following
any spillage of radioisotopes. Laboratory work should be performed in
stainless steel trays lined with absorbent paper.

::\*Excretion of radioactive materials by patients may be via faeces,
urine, saliva, blood, exhaled breath or the skin. Provision to deal with
any or all of these potential pathways for contamination must be made.

::\*Provision for collection and possible storage of both liquid and
solid radioactive waste may be necessary in some circumstances. Most
short-lived, water soluble liquid waste can be flushed into the sewers
but longer lived isotopes such as I-131 may have to be stored for decay.
Such waste must be adequately contained and labelled during storage.
`<b>`{=html}Equipment`</b>`{=html}

::\*Ensure that appropriate survey monitors are available to determine
if any contamination has occurred and to assist in decontamination
procedures. Routine monitoring of potentially contaminated areas must be
performed.

::\*Ensure that all potentially exposed staff are issued with individual
personnel monitors.

::\*Protective clothing such as gowns, smocks, overboots and gloves
should be provided and worn to prevent contamination of the personnel
handling the radioactivity. In particular, gloves must be worn when
administering radioactive materials orally or intravenously to patients.
It should be noted that penetration of gloves may occur when handling
some iodine compounds so that wearing a second pair of gloves is
recommended. In any event, gloves should be changed frequently and
discarded ones treated as radioactive waste.
`<b>`{=html}Behaviours`</b>`{=html}

::\*Eating and drinking of food, smoking, and the application of
cosmetics is prohibited in laboratories in which unsealed sources are
utilized.

::\*Mouth pipetting of any radioactive substance is totally prohibited.

::\*Precautions should be taken to avoid punctures, cuts, abrasions and
any other open skin wounds which otherwise might allow egress of
radiopharmaceuticals into the blood stream.
`<b>`{=html}Optimization`</b>`{=html}

::\*Always ensure that there is a net benefit resulting from the patient
procedure. Can the diagnosis or treatment be made by recourse to an
alternative means using non ionizing radiation?

::\*Ensure that all staff, including physicians, technologists, nurses
and interns and other students, who are involved in the practice of
nuclear medicine receive the relevant level of training and education
appropriate to their assigned tasks. The training program could be in
the form of seminars, refresher courses and informal tutorials.

::\*A substantive Quality Assurance (QA) program should be implemented
to ensure that the function of the Dose Calibrator, Gamma Camera,
computer and other ancillary equipment is optimized.

The potential hazards to staff in a nuclear medicine environment
include:

::\*Milking the ^99m^Tc generator, drawing up and measuring the quantity
of radioisotope prior to administration.

::\*Delivering the activity to the patient by injection or other means
and positioning the now radioactive patient in the imaging device.

::\*Removing the patients from the imaging device and returning them to
the ward where they may continue to represent a radiation hazard for
some time. For Tc-99m, a short-lived radionuclide the hazard period will
be only a few hours but for therapeutic isotopes the hazardous period
may be several days.

::\*Disposal of radioactive waste including body fluids, such as blood
and urine, but also swabs, syringes, needles, paper towels etc.

::\*Cleaning up the imaging area after the procedure.

::\*Contamination.

The table below lists the dose rates from patients having nuclear
medicine examinations. In general, the hazards from handling or dealing
with radioactive patients arise in two parts:

::\*External hazard: This will be the case when the radioisotope emits
penetrating gamma-rays. Usually, this hazard can be minimised by
employing shielding and sensible work practices.

::\*Radioactive contamination: This is potentially of more concern as it
may lead to the inhalation or ingestion of radioactive material by
staff. Possible sources of contamination are radioactive blood, urine
and saliva, emanating from a patient, or airborne radioactive vapour.
Sensible work practices, which involve high levels of personal hygiene,
should ensure that contamination is not a major issue.

!Dose rates from patients following
administration.{width="600"}

One of the most common nuclear medicine diagnostic procedures is the
bone scan using the isotope Tc-99m. The exposure rate at 1 metre from a
typical patient will peak at approximately 3 μSv per hour immediately
after injection dropping steadily because of radioactivity decay and
through excretion so that after 2 hours it will be about 1.5 μSv per
hour. Neglecting any further excretion, the total exposure received by
an individual, should that person stand one meter from the patient for
the whole of the first 24 hours, would be \~17 μSv. For a person at 3
meters from the patient this number would reduce to 1.7 μSv and for a
distance of 5 metres it would be \~0.7 μSv. These values have been
estimated on the basis of the inverse square law.

Patients should be encouraged to drink substantial quantities of liquid
following their scan, as this will improve excretion and aid in
minimizing not only their radiation dose but also that of nursing staff.

![](radSafetyNutShell.jpg "radSafetyNutShell.jpg"){width="480"}


# 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*.

```{=html}
<div class="center">
```
![](NM7_1.gif "NM7_1.gif"){width="320"}

```{=html}
</div>
```
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

```{=html}
<div class="center">
```
$I_x < I_0\,\!$

```{=html}
</div>
```
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

```{=html}
<div class="center">
```
$\Delta I = I_0 - I_x\,\!$

```{=html}
</div>
```
#### 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).

```{=html}
<!-- -->
```

:   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.

```{=html}
<!-- -->
```

:   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.

```{=html}
<!-- -->
```

:   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.

```{=html}
<!-- -->
```

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

```{=html}
<div class="center">
```
$\Delta I \propto Z^3\,\!$

```{=html}
</div>
```

:   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.

```{=html}
<!-- -->
```

:   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.

```{=html}
<div class="center">
```
![](NM7_6.gif "NM7_6.gif"){width="220"}

```{=html}
</div>
```

:   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:

```{=html}
<div class="center">
```
$-d I \propto I \cdot dx\,\!$

```{=html}
</div>
```
the minus sign indicating that the intensity is reduced by the absorber.

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

```{=html}
<div class="center">
```
$-d I = \mu I \cdot dx\,\!$

```{=html}
</div>
```
where the constant of proportionality, μ, is called the Linear
Attenuation Coefficient.

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

```{=html}
<div class="center">
```
$-\frac{d I}{I} = \mu \cdot dx$

```{=html}
</div>
```
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:

```{=html}
<div class="center">
```
$- \int_{I_0}^{I_x} \frac{dI}{I} = \mu \int_{0}^x dx$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \ln \left ( \frac{I_x}{I_0} \right ) = -\mu x$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \frac{I_x}{I_0} = \text{exp}\ (-\mu x)$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore I_x = I_0\ \text{exp}\ (-\mu x)\,\!$

```{=html}
</div>
```
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.

```{=html}
<div class="center">
```
  ------------------------------------------- -------------------------------------------
  ![](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.

```{=html}
</div>
```
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:

```{=html}
<div class="center">
```
$I_x = \frac{I_0}{2}$

```{=html}
</div>
```
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:

```{=html}
<div class="center">
```
$I = \frac{I_0}{2}$

```{=html}
</div>
```
when

```{=html}
<div class="center">
```
$x = x_{\frac{1}{2}}$

```{=html}
</div>
```
and inserting it in the exponential attenuation equation, that is:

```{=html}
<div class="center">
```
$I = I_0\ \text{exp}\ (-\mu x)\,\!$

```{=html}
</div>
```
to give

```{=html}
<div class="center">
```
$\frac{I_0}{2} = I_0\ \text{exp}\ (-\mu x_{\frac{1}{2}})$

```{=html}
</div>
```
Therefore

```{=html}
<div class="center">
```
$\frac{1}{2} = \text{exp}\ (-\mu x_{\frac{1}{2}})$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore 2^{-1} = \text{exp}\ (-\mu x_{\frac{1}{2}})$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \ln 2^{-1} = -\mu x_{\frac{1}{2}}$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore \ln 2 = \mu x_{\frac{1}{2}}$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore 0.693 = \mu x_{\frac{1}{2}}$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\mu = \frac{0.693}{x_{\frac{1}{2}}}$

```{=html}
</div>
```
and

```{=html}
<div class="center">
```
$x_{\frac{1}{2}} = \frac{0.693}{\mu}$

```{=html}
</div>
```
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:

```{=html}
<div class="center">
```
$\frac{\text{Linear Attenuation Coefficient}}{\text{Density}} = \frac{\mu}{\rho}$

```{=html}
</div>
```
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:

```{=html}
<div class="center">
```
$I = \frac{I_0}{10},\ \text{when}\ x = \text{?}$

```{=html}
</div>
```

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

```{=html}
<div class="center">
```
$\mu = \frac{0.693}{x_{\frac{1}{2}}} = \frac{0.693}{2.14} = 0.324\ \text{cm}^{-1}$

```{=html}
</div>
```

:   Now combining all this with the exponential attenuation equation:

```{=html}
<div class="center">
```
$I = I_0\ \text{exp}\ (-\mu x)\,\!$

```{=html}
</div>
```

:   we can write:

```{=html}
<div class="center">
```
$\frac{I_0}{10} = I_0\ \text{exp}\ (-0.324x)$

```{=html}
</div>
```

:   Therefore

```{=html}
<div class="center">
```
$\frac{1}{10} = \text{exp}\ (-0.324x)$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore -\ln 10 = -0.324x$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore x = \frac{\ln 10}{0.324} = \frac{2.3}{0.324} = 7.1\ \text{cm}$

```{=html}
</div>
```
```{=html}
<div class="center">
```
$\therefore x \approx 7\ \text{cm}$

```{=html}
</div>
```

:   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.

```{=html}
<!-- -->
```

:   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!

```{=html}
<!-- -->
```

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

```{=html}
<!-- -->
```

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

```{=html}
<!-- -->
```

:   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!

```{=html}
<!-- -->
```

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

```{=html}
<div class="center">
```
3.3 R hr^-1^ mCi^-1^ at 1 cm from the source.

```{=html}
</div>
```

:   This is equal to:

```{=html}
<div class="center">
```
3300 mR hr^-1^ mCi^-1^ at 1 cm from the source,

```{=html}
</div>
```

:   which is equal to:

```{=html}
<div class="center">
```
$\frac{3300}{(100)^2}\ \text{mR hr}^{-1}\ \text{mCi}^{-1}\ \text{at 1 m from the source}$

```{=html}
</div>
```

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

```{=html}
<div class="center">
```
$\frac{3300}{10^4 (3.7 \cdot 10^7)}\ \text{mR hr}^{-1}\ \text{Bq}^{-1}\ \text{at 1 m from the source}$

```{=html}
</div>
```

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

```{=html}
<div class="center">
```
$\frac{3300 \cdot 10^5 \cdot 10^6}{10^4 (3.7 \cdot 10^7)}\ \text{mR hr}^{-1}\ (10^5\ \text{MBq})^{-1}\ \text{at 1 m from the source}$

```{=html}
</div>
```

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

```{=html}
<!-- -->
```

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

```{=html}
<!-- -->
```

:   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.