# 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!

