11.3: Radioactivity

Learning Objectives

1. Describe the model for the number of radioactive atoms and explain how this leads to a differential equation.
2. Determine the solution of the resulting differential equation.
3. Given the initial amount, determine the amount of radioactivity remaining at a future time.
4. Describe the link between half-life of the radioactive material and its decay rate; given the value of one, be able to find the value of the other.

A radioactive material consists of atoms that undergo a spontaneous change. Every so often, some radioactive atom emits a particle, and decays into an inert form. We call this a process of radioactive decay. For any one atom, it is impossible to predict when this event would occur exactly, but based on the behavior of a large number of atoms decaying spontaneously, we can assign a probability \(k\) of decay per unit time.

In this section, we use the same kind of book-keeping (keeping track of the number of radioactive atoms remaining) as in the population growth example, to arrive at a differential equation that describes the process. Once we have the equation, we determine its solution and make a long-term prediction about the amount of radioactivity remaining at a future time.

Deriving the model

We start by letting \(N(t)\) be the number of radioactive atoms at time \(t\). Generally, we would know \(N(0)\), the number present initially. Our goal is to make simple assumptions about the process of decay that allows us to arrive at a mathematical model to predict values of \(N(t)\) at any later time \(t>0\).

Assumptions.

1. The process of radioactive decay is random, but on average, the probability of decay for a given radioactive atom is \(k\) per unit time where \(k>0\) is some constant.
2. During each (small) time interval of length \(\Delta t=h\), a radioactive atom has probability \(k h\) of decaying. This is merely a restatement of (1).

Suppose that at some time \(t\), there are \(N(t)\) radioactive atoms. Then, according to our assumptions, during the time period \(t \leq t \leq t+h\), on average \(k h N\left(t_{0}\right)\) atoms would decay. How many are there at time \(t+h\) ? We can write the following balance-equation:

\[\left[\begin{array}{c} \text { Amount left } \\ \text { at time } \\ t+h \end{array}\right]=\left[\begin{array}{c} \text { Amount present } \\ \text { at time } \\ t \end{array}\right]-\left[\begin{array}{c} \text { Amount decayed } \\ \text { during time interval } \\ t \leq t \leq t+h \end{array}\right] \nonumber \]

or, restated:

\[N(t+h)=N(t)-k h N(t) \]

Here we have assumed that \(h\) is a small time period. Rearranging Equation (11.3.1) leads to

\[\frac{N(t+h)-N(t)}{h}=-k N(t) . \nonumber \]

Considering the left hand side of this equation, we let \(h\) get smaller and smaller \((h \rightarrow 0)\) and recall that

\[\lim _{h \rightarrow 0} \frac{N(t+h)-N(t)}{h}=\frac{d N}{d t}=N^{\prime}(t) \nonumber \]

where we have used the notation for a derivative of \(N\) with respect to \(t\). We have thus shown that a description of the population of radioactive atoms reduces to

\[\frac{d N}{d t}=-k N \]

We have, once more, arrived at a differential equation that provides a link between a function of time \(N(t)\) and its own rate of change \(d N / d t\). Indeed, this equation specifies that \(d N / d t\) is proportional to \(N\), but with a negative constant of proportionality which implies decay.

Above we formulated the entire model in terms of the number of radioactive atoms. However, as shown below, the same equation holds regardless of the system of units used measure the amount of radioactivity

Mastered Material Check

24. Suppose a given atom has a \(1 \%\) chance of decay per 24 hours. What is this atom’s probability of decay per week? Per hour?

Example 11.7

Define the number of moles of radioactive material by \(y(t)=\) \(N(t)\) / \(A\) where \(A\) is Avogadro’s number (the number of molecules in 1 mole: \(\approx 6.022 \times 10^{23}\) - a dimensionless quantity, i.e. just a number with no associated units). Determine the differential equation satisfied by \(y(t)\).

Solution

We write \(y(t)=N(t) / A\) in the form \(N(t)=A y(t)\) and substitute this expression for \(N(t)\) in Equation (11.9). We use the fact that \(A\) is a constant to simplify the derivative. Then

\[\frac{d N}{d t}=-k N \quad \Rightarrow \quad \frac{A d y(t)}{d t}=-k(A y(t)) \quad \Rightarrow \quad A \frac{d y(t)}{d t}=A(-k y(t)) \nonumber \]

cancelling the constant \(A\) from both sides of the equations leads to

\[\frac{d y(t)}{d t}=-k y(t) \nonumber \]

or simply

\[\frac{d y}{d t}=-k y \]

Thus \(y(t)\) satisfies the same kind of differential equation (with the same negative proportionality constant) between the derivative and the original function. We will refer to (11.3.3) as the decay equation.

Solution to the decay equation (11.3.3)

Suppose that initially, there was an amount \(y_{0}\). Then, together, the differential equation and initial condition are

\[\frac{d y}{d t}=-k y, \quad y(0)=y_{0} \]

We often refer to this pairing between a differential equation and an initial condition as an initial value problem. Next, we show that an exponential function is an appropriate solution to this problem

Example 11.8 (Checking a solution)

Show that the function

\[y(t)=y_{0} e^{-k t} \]

is a solution to initial value problem (11.3.4).

Solution

Ee compute the derivative of the candidate function (11.3.5), and rearrange, obtaining

\[\frac{d y(t)}{d t}=\frac{d}{d t}\left[y_{0} e^{-k t}\right]=y_{0} \frac{d e^{-k t}}{d t}=-k y_{0} e^{-k t}=-k y(t) . \nonumber \]

This verifies that for the derivative of the function is \(-k\) times the original function, so satisfies the DE in (11.11). We can also check that the initial condition is satisfied:

\[y(0)=y_{0} e^{-k \cdot 0}=y_{0} e^{0}=y_{0} \cdot 1=y_{0} . \nonumber \]

Hence, Equation (11.3.5) is the solution to the initial value problem for radioactive decay. For \(k>0\) a constant, this is a decreasing function of time that we refer to as exponential decay.

The half life

Given a process of exponential decay, how long would it take for half of the original amount to remain? Let us recall that the "original amount" (at time \(t=0\) ) is \(y_{0}\). Then we are looking for the time \(t\) such that \(y_{0} / 2\) remains. We must solve for \(t\) in

\[y(t)=\frac{y_{0}}{2} . \nonumber \]

We refer to the value of \(t\) that satisfies this as the half life.

Example 11.9 (Half life)

Determine the half life in the exponential decay described by Equation (11.12).

Solution

We compute:

\[\frac{y_{0}}{2}=y_{0} e^{-k t} \quad \Rightarrow \quad \frac{1}{2}=e^{-k t} . \nonumber \]

Now taking reciprocals:

\[2=\frac{1}{e^{-k t}}=e^{k t} . \nonumber \]

Thus we find the same result as in our calculation for doubling times, namely,

\[\ln (2)=\ln \left(e^{k t}\right)=k t, \nonumber \]

so that the half life is

\[\tau=\frac{\ln (2)}{k} . \nonumber \]

This is shown in Figure 11.7.

Figure 11.7: Half-life in an exponentially decreasing process.

Example 11.10

(Chernobyl: April 1986) In 1986 the Chernobyl nuclear power plant exploded, and scattered radioactive material over Europe. The radioactive element iodine-131 \(\left(I^{131}\right)\) has half-life of 8 days whereas cesium\(137\left(\mathrm{Cs}^{137}\right)\) has half life of 30 years. Use the model for radioactive decay to predict how much of this material would remain over time.

Solution

We first determine the decay constants for each of these two elements, by noting that

\[k=\frac{\ln (2)}{\tau}, \nonumber \]

and recalling that \(\ln (2) \approx 0.693\). Then for \(\mathrm{I}^{131}\) we have

\[k=\frac{\ln (2)}{\tau}=\frac{\ln (2)}{8}=0.0866 \text { per day } \nonumber \]

Then the amount of \(\mathrm{I}^{131}\) left at time \(t\) (in days) would be

\[y_{I}(t)=y_{0} e^{-0.0866 t} . \nonumber \]

For \(\mathrm{Cs}^{137}\)

\[k=\frac{\ln (2)}{30}=0.023 \text { per year. } \nonumber \]

so that for \(T\) in years,

\[y_{C}(T)=y_{0} e^{-0.023 T} . \nonumber \]

Note: we have used \(T\) rather than \(t\) to emphasize that units are different in the two calculations done in this example.

Example 11.11 (Decay to \(0.1\%\) of the initial level)

How long it would take for \(I^{131}\) to decay to \(0.1 \%\) of its initial level? Assume that the initial level occurred just after the explosion at Chernobyl.

Solution

We must calculate the time \(t\) such that \(y_{I}=0.001 y_{0}\) :

\(0.001 y_{0}=y_{0} e^{-0.0866 t} \Rightarrow 0.001=e^{-0.0866 t} \Rightarrow \ln (0.001)=-0.0866 t\).

Therefore,

\[t=\frac{\ln (0.001)}{-0.0866}=\frac{-6.9}{-0.0866}=79.7 \text { days. } \nonumber \]

Thus it would take about 80 days for the level of Iodine- 131 to decay to \(0.1 \%\) of its initial level.

Exercise \(\PageIndex{1}\)

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Answer

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Mastered Material Check

25. Repeat the calculation in Example 11.11 for Cesium.
26. Convert the Cesium decay time units to days and repeat the calculation of Example 11.10 with the new time units.
27. If the decay rate of a substance is 10% per day, what is its half-life?