Objectives:

• To understand and relate to the simulation of radioactive decay using dice
• To understand the terms; ‘activity’ of a source and ‘decay constant’ of a radioisotope
• To be able to understand and use the equation for activity if a radioisotope; $A = \lambda N$
• To be able to use graphical methods and spreadsheet modelling of the equation $\frac{\Delta N}{\Delta t} = - \lambda N$ for radioactive decay.
• To be able to understand and use the equations for determining the activity or number of undecayed nucle remaining of a radioactive substance; $A = A_{0} e^{- \lambda t}$ and  $N = N_{0} e^{- \lambda t}$

Radioactive decay is the process by which unstable nuclei lose energy by emitting a form radioactive radiation. There are three main types of nuclear radiation that have been discussed previously; alpha ($\alpha$), beta ($\beta$) and gamma ($\gamma$) radiation. Radioactive decay is a spontaneous process whereby a single unstable nuclei will emit either, $\alpha$ $\beta$ or $\gamma$ radiation to become more stable.

As discussed on the page explaining $\beta ^{-}$ and $\beta ^{+}$ decay, if there is an incorrect ratio of the number protons to the number of neutrons it is likely that the radioisotope will undergo a form of beta decay;

• Excess neutrons – $\beta^{-}$ decay:  $d \ \ \rightarrow \ \ u + \ ^{0}_{-1}e + \ \bar{\nu_{e}}$
• Excess protons $\beta^{+}$ decay: $u \ \ \rightarrow \ \ d + \ ^{0}_{+1}e + \ {\nu_{e}}$

The type predicted using the Segrè plot:

If a radioisotope can become more stable by emitting an alpha particle (the reason for why this is possible will be given after studying ‘binding energy’) then it will do this instead, gamma radiation can be emitted due to an atom in an excited state becoming more de-excited or a by-product in another decay such as fission.

It is impossible to predict when any individual nucleus will undergo radioactive decay due to its spontaneous nature. However, the rate of decay of a nucleus in a particular sam

ple can be predicted and therefore statistics can be used to estimate how many will decay in a period of time.

In class it is likely you will carry out the following experiment which simulates radioactive decay and can then help the terms ‘activity’ and ‘decay constant’ to be understood;

Imagine a collection of cubes all painted the same colour, say yellow, but one side of each and every cube is painted a different colour, black. Rolling a single die gives a one in six chance of the black side landing face up, if a die lands on the black side we will say that it has ‘decayed’ – hence the link to radioactivity. The way this links to radioactive decay is that, for any one particular roll it is impossible (for an unbiased die) to predict whether the die lands on the black side and decays or not – statistically there is a one in six chance but this probability will reset after any individual roll.

Having a selection of lets say 50 dice (over a hundred is needed for any good results to be seen) will enable you to simulate the decay of a radioactive sample that has 50 undecayed nuclei. At time t = 0, we can say that none of the dice have ‘decayed’, so we would have 50 dice remaining. We can simulate time by rolling all of the dice, each roll of the set can represent one time interval. The following table can then be drawn;

After each roll, the number of dice can refining should be counted and the number out into the table. So lets say, 7 decayed in the first roll and 8 in the second, 3 in the third, 0 in the fourth and 5 in the fifth, the table would become;

Continuing this until all the dice have ‘decayed’ and then plotting a graph of dice remaining against the roll number will give a graph that may appear like;

The individual plots show the spontaneous nature of each roll (the amount decaying each time varying). The line of best fit tends towards zero which should be expected because you would keep rolling until you have none left.

The result above is quite can vary, below are two more examples, If you look closely at the data points you can see the variability of the data from such an experiment;

All of the graphs above show an exponential trend but to justify this more dice should be used (over 100 was mentioned above). Below is another graph collating 8 sets of data and so showing the relation of 400 dice ‘decaying’. The result is a much smoother set of data with a trend line following it much more closely;

Using this graph, what factors affect the rate in which dice would decay?

There are two factors;

1. Each die had just one side that was painted black. What if two sides were painted black, or three? Increasing the number of sides would increase the chance of each dice decaying and therefore would increase the rate of decay and as such would give the graph a steeper gradient.
2. In this example we mentioned one time interval being equal to one roll, well what if a time interval were to equal two rolls, or three? This would mean that for each time interval, the number of dice that decay would increase.

In reality, there are no sides to nuclei that decay and we don’t roll anything to make them decay. Instead these are properties of a radioisotope known as the activity and the decay constant.

The decay constant, denoted with the letter $\lambda$, is (as the term suggests) a constant. In comparing it to the dice simulation the decay constant would be the likelihood for any individual dice to decay, so 1 in 6 or $= 0.1666667$ . This value is constant no matter how many dice are remaining.

The activity, denoted by the letter $A$, is the rate at which a radioactive source decays. When measuring the radioactivity of a substance you would measure the activity of it, this is done using a Geiger-Müller tube and counter – click here for a reminder. This is different tow the decay constant because it depends on how many of undecayed nuclei there are at any given moment in time. As the graphs above show, there is a greater rate of decay when there are more undecayed nuclei. When there is only one undecayed nuclei remaining, this may take a second to decay, maybe an hour, maybe even 1000 years. Therefore the activity is changing. The activity to commonly measured in the number of decays (or iterations) per second, and so had a unit of $s^{-1}$ .

Activity and decay constant link depending on the number of undecayed nuclei by the formula;

$A = \lambda N$     (1)

Since the decay constant is a probability for an undecayed nuclei to decay, it makes sense that it should always be less than or equal to 1 and therefore the activity can never be greater that the number of undecayed nuclei remaining.

The activity of a radioisotope can also be found using the formula;

$A = -\frac{\Delta N}{\Delta t}$     (2)

This is the rate at which the undecayed nuclei decay. There is a negative in front of the $latex \frac{\Delta N}{\Delta t}$ because as the sample decays, there are fewer undecayed nuclei remaining and therefore the activity decreases. The negative account of the decreasing change in activity.

Equatiing equations (1) and (2) gives;

$\frac{\Delta N}{\Delta t} = - \lambda N$     (3)

From this graphical methods and spreadsheet modelling can be carried to determine the significance of the decay constant and activity, when using radioactive materials.

By rearranging, multiplying, integrating, taking logs and further rearrangement of equation (3) an equation for how the activity changes over time can be found;

$A = A_{0}e^{- \lambda t}$

A full derivation can be found on the page titled ‘Deriving the Equation for the Changing Activity of a Radioactive Substance‘.

By dividing through the decay constant an equation for how the number of undecayed nuclei changes over time can be found;

$N = N_{0}e^{- \lambda t}$

A full derivation can be found on the page given above as well.

Example of using the  $A = \lambda N$  equation;

Imagine we have 1000 undecayed nuclei of Radon (unrealistic, but just go with it),

The decay constant for radon is  $\lambda_{radon} = 1.33 \times 10^{-2} s^{-1}$

Using  $A = \lambda N$  we can see that;

$A = 1.33 \times 10^{-2} \times 1000 = 13.3 s^{-1}$

This tells us that when there are 100 undecayed nuclei remaining they will decay at a rate of approximately 13.3  per second – However, don’t forget that that radioactivity is spontaneous and that this is just the statistical likelihood for decay.

Example of using the  $N(t) = N_{0} e^{(-\lambda t)}$  equation;

Approximately how many undecayed nuclei will still be remaining after 52 seconds?

$N_{0} = 1000$
$\lambda = 1.33 \times 10^{-2} s^{-1}$
$t = 52 s$
$N = ?$

$N(t) = N_{0} e^{(-\lambda t)}$

$N(t) = 1000 e^{(-1.33 \times 10^{-2} \times 52)}$

$N(t) = 500.77$

So $N$ has approximately halved $(\frac{500}{1000})$

Taking this sample of radon but after 52 seconds, will the activity of this substance have changed?

Using  $A = \lambda N$  we can see that;

$A = 1.33 \times 10^{-2} \times 500$

$A = 6.65 s^{-1}$

So;

$A_{N = 1000)} = 13.3 s^{-1}$
$A_{N = 500)} = 6.65 s^{-1}$

These values tell us that the activity has reduced, since it is dependent on the number of undecayed nuclei. However, the decay constant, $\lambda$  is constant.

This same outcome can be proved by using  $A(t) = A_{0} e^{(-\lambda t)}$ ;

$A_{0} = 13.3 s^{-1}$
$\lambda = 1.33 \times 10^{-2} s^{-1}$
$t = 52 s$
$A = ?$

$A(t) = A_{0} e^{(-\lambda t)}$

$A(t) = 13.3 e^{(-1.33 \times 10^{-2} \times 52)}$

$A(t) = 6.66 \approx 6.65 s^{-1}$

and so it compares to the previous answer.