Radioactive isotopes have the curious property that in time they break down into other, more stable, substances. This process is known as radioactive decay. While there is no way to predict when an individual radioactive atom will decay, the decay of a large collection of radioactive isotopes is very regular and can be predicted.
For example, if we have two pounds of the radioactive isotope Cesium-137 \((\text{Cs}{137})\) it is known that after the passage of \(30\) years and \(70\) days we will have only one pound. The rest will have decayed into barium. Even more curious though, is this: If we wait another \(30\) years and \(70\) days only one–half pound of the remaining one pound will have decayed. The numbers change depending on the isotope but the general principle is that a fixed percentage of a radioactive isotope will decay during a fixed interval of time.
Do you see the implications of this? Since a fixed percentage decays in a fixed time interval the rate of change of \(A(t)\) will be proportional to \(A\) itself. Since this is exactly what we said about both population growth and return on investment, IVP (8.19) should model this phenomenon as well. There is, of course, the difference that the quantity is now shrinking rather than growing. How do you think that will manifest in our model? Take a guess.
Example8.10.1.
For the sake of being definite suppose that our isotope is decaying (shrinking) at a relative (nominal) rate of \(25\%\text{.}\) Then if we start with \(20\)Kg, the IVP to be solved is
Wait a minute! This can’t be right. It says that the number of isotopes is increasing, not decreasing. Graph it and see.
So what did we do wrong? What does your intuition say?
The derivative of \(A(t)\) is its rate of change. If \(A(t)\) is increasing then its rate of change is going to be positive. If it is decreasing then its rate of change will be negative. We seem to have committed one of the classic blunders of mathematics: We got the sign wrong. We should have been solving the IVP
Graph \(A(t)=20e^{-0.25t}\) to confirm that it is decreasing.
Did you guess correctly that the sign of the exponent would be negative in this model?
Example8.10.3.
Suppose we have \(6\) pounds of a very unstable isotope which is known to lose half of its mass to radioactive decay in \(2\) hours. If we let \(y(t)\) be the mass of our isotope sample at any time \(t\text{,}\) then the IVP we need to solve is
As always the solution of a differential equation of this form is
\begin{equation*}
y(t) =6e^{rt}
\end{equation*}
where and \(r\) is an unknown constant. To find \(r\) we must use the additional information given. Half of our isotope’s mass is lost in \(2\) hours, so \(y(2)=3\text{.}\) Thus
This is correct but again it is cumbersome. Also the lack of an obvious negative in the exponent is a little troubling.
Problem8.10.4.
Use the properties of logarithms and exponentials to show that \(6e^{\frac{t}{2}\ln\left(\frac12\right)} =
6e^{-\frac{t}{2}\ln\left(2\right)}\text{.}\)
Use the properties of logarithms and exponentials to show that \(6e^{\frac{t}{2}\ln\left(\frac12\right)} = \frac{6}{\left(\sqrt{2}\right)^t}\)
The length of time that it takes for a quantity of a radioactive isotope to decay by one-half is called the half–life of the isotope. Each substance has its own half-life. The half–life for Carbon-14 is roughly \(5,730\) years; for Fluorine-18, it is \(110\) minutes; for Potassium–\(40\text{,}\) it is \(1.25\) billion years.
Problem8.10.5.
A certain isotope has a half–life of \(h\text{,}\) and is decaying exponentially:
Use the result of part (a) to show that the amount of our radioactive material at time \(t\) is given by \(A(t)=A_0 e^{\frac{1}{h} \ln\left(\frac{1}{2}\right)t}.\)
(c)
The formula in part (a) is correct but it is a little awkward to use, as written. Use the properties of exponents and logarithms to show that this can be rewritten as \(A(t)=A_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}. \) What are \(A(0),\)\(A(h),\)\(A(2h),\) and \(A(3h)?\) Is this consistent with calling \(h\) the half–life? Explain.
Radiocarbon \((C_{14})\) is a radioactive isotope of carbon which is constantly being created in the atmosphere by the interaction of cosmic rays with atmospheric nitrogen. The resulting radiocarbon combines with atmospheric oxygen to form radioactive carbon dioxide, which is incorporated into plants by photosynthesis. Animals then acquire \(C_{14}\) by eating the plants. When the animal or plant dies, it stops absorbing carbon from the environment. From that point onward the amount of \(C_{14}\) it contains begins to decrease as the \(C_{14}\) undergoes radioactive decay. During life the ratio of \(C_{14}\) to other, stable, isotopes remains constant (more or less). However after an organism dies this ratio begins to decrease due to the decay of \(C_{14}\)
By measuring the amount of \(C_{14}\) in a sample from a dead plant or animal such as a piece of wood or a fragment of bone we can calculate, approximately when the plant or animal died. This is called radiocarbon dating, and it was developed by Willard Libby 12
(1908–1980), in \(1949\text{.}\) Radiocarbon dating revolutionized the field of archaeology and Libby was awarded the Nobel Prize in Chemistry in \(1960\) for this development.
Problem8.10.6.
The Shroud of Turin, shown below, is a Christian religious relic which bears an image of a man. Some people believe it is the burial cloth of Jesus and that the image is that of Jesus himself. In \(1987\) the Vatican agreed to subject pieces of the shroud to radiocarbon dating. In this problem we will recreate the computations done to determine the age of the Shroud. Let \(A=A(t)\) be the amount of \(C_{14}\) (in mg) at time \(t\) years, where \(t=0\) represents when the shroud was used. Let \(A(0)=A_0\) be the initial amount of \(C_{14}\) present in the sample.
(a)
How much \(C_{14}\) would be present if the shroud was 2000 years old?
(b)
If one of the samples of the Shroud contained 88.9% of the original \(C_{14}\text{.}\) How old would this sample be?
In medicine, Positron Emission Tomography (PET) scans use radioactive tracers to image body functions. One of the most commonly used radioactive tracers is Fluorine–\(18\)\((F_{18})\) which has a half life of \(110\) minutes. Typically \(F_{18}\) is injected into the body and the imaging is done about one hour after the tracer is injected. Suppose that \(s\) units of \(F_{18}\) must remain in the body for the PET scan results to be useful. How much \(F_{18}\) must be injected into the patient \(60\) minutes prior?
Drill8.10.8.
The half–life of \(C_{14}\) limits it to dating artifacts that are no older than \(50,000\) years. What percentage of the original amount of \(C_{14}\) would remain in a \(50,000\)–year-old artifact? How does this explain the limitation on radiocarbon dating?
Problem8.10.9.
The radioactive isotope Potassium–\(40\)\((K_{40})\) has a half life of \(1.25\) billion years. Radiometric dating using \(K_{40}\) is especially effective for dating very old volcanic rock as the quickly cooling lava traps the Argon formed by the decaying of \(K_{40}.\) This has been used by scientists to study the frequency of geomagnetic reversals. A geomagnetic reversal is a change in the earth’s magnetic polarity where the magnetic north and south poles (not to be confused with the geographic north and south poles) are switched. The latest such reversal is called the Brunhes-Matuyama 14
reversal. Basically, the magnetic polarity of the planet is “recorded” in cooled lava flows. By dating the age of the lava flows, scientists can date these reversals. There are limitations to this dating method, as the smallest percentage of \(K_{40}\) that can be detected is about \(0.0053\%\text{.}\)
(a)
According to the Wikipedia article we referenced above the the Brunhes–Matuyama reversal occured approximately \(781,000\) years ago. A rock sample which appears to have a magnetic reversal contains between \(99.9541\%\) and \(99.9593\%\) of the \(K_{40}\) that it originally contained. Is this consistent with the Brunhes–Matuyama reversal?
(b)
Approximately what is the age of the youngest rock that can be dated using this technique?
Subsection8.10.2Chillin’ with Newton: The Law of Cooling
Example8.10.10.
Suppose a container of water at a temperature of \(40^\circ F\) is placed into a freezer maintained at a constant temperature of \(5^\circ F.\) We’d like to have a model that will allow us to compute how long it will take for the water to cool down to a temperature of, for example, \(32^\circ F\) where it will start to freeze.
In the late \(17\)th century Isaac Newton showed experimentally that the rate at which the water cools will be proportional to the difference between its current temperature and the ambient (surrounding) temperature. This is known as Newton’s Law of Cooling and it holds generally, not just for water.
For this problem we see that the ambient temperature is \(5^\circ F\) and the initial temperature of the water is \(40^\circ F\text{.}\) We let \(T(t)\) represent the temperature of the water at any given time. Then the difference between the current temperature and the ambient temperature is \(T-5\text{,}\) and (as always) the rate of change of the temperature of the water with respect to time is \(\dfdx{T}{t}\text{,}\) (or \(\dot{T}\) if you are Newton). Letting \(r\) be the (unknown) constant of proportionality we have the IVP
This differential equation looks like it will be tougher to solve than the IVPs we dealt with earlier. But in fact we can make this problem “easier on the eyes” with the substitution \(D=T-5\text{.}\) Notice that \(\dfdx{D}{t}=\dfdx{(T-5)}{t}=\dfdx{T}{t}\) so we have the IVP
To complete our model we need to determine \(r\text{.}\) We can find \(r\) if we know the temperature of the water at a second time during the cooling. Assume that we measured the temperature of the water after \(10\) minutes and it was \(35^\circ F.\) Then we have
Newton’s Law of Cooling accurately models the cooling of any object placed in cooler surroundings. We used water in our example just to be definite.
Problem8.10.11.
Since the temperature of the water is decreasing we would expect \(r\) to be less than zero. Show that it is.
Problem8.10.12.
(a)
Answer the original question: How long does it take for the water to cool from \(40^\circ F\) to \(32^\circ F\)
(b)
Take a guess: Would it take the same amount of time to cool another \(8^\circ F\) from \(32^\circ F\) to \(24^\circ F?\) Check to see if your guess was correct.
(c)
What would the initial temperature of the water need to be to take twice as long to cool to \(32^\circ F\) as it did for the \(40^\circ F\) water?
Problem8.10.13.
The rule of thumb forensic scientist use to determine at person’s the time of death is to start with a body temperature of \(37^\circ C\) and subtract \(1.5^\circ C\) for each hour the person is dead. Of course, this simple linear model isn’t as accurate as Newton’s Law of Cooling since it does not take into account the surrounding temperature, size of the body, etc. and can only be applied until the temperature of the body reaches the surrounding temperature.
(a)
Suppose the ambient temperature is \(20^\circ C.\) Using the forensic rule of thumb, how long would it take for the body to reach the ambient temperature? Using the forensic rule of thumb, what would the body temperature be halfway through the cooling process.
(b)
Use the temperature predicted by the rule of thumb at the halfway point to give a complete model of the body’s cooling as predicted by Newton’s Law of Cooling.
(c)
Of course, in Newton’s Law of Cooling, the body temperature will never quite reach the ambient temperature, but just for comparison, substitute the time when the rule of thumb predicts that the body temperature is \(20^\circ C\) and see how close it is to the ambient temperature.
(d)
Using the answer you obtained in part (c), find the rate of change of the body temperature at the beginning of the time interval and the rate of change of the body temperature at the end of the time interval. How do these compare with the rule of thumb rate of change?
Of course, Newton’s Law of Cooling can be used just as well to model an object heating. Consider the following:
Problem8.10.14.
A whole turkey is considered to be safely cooked when the internal temperature is \(165^\circ F.\) Suppose a turkey is taken out of a refrigerator set at \(35^\circ F\) and is put directly into an oven set at \(325^\circ F\text{.}\) You check it \(2\) hours later and the internal temperature is \(100^\circ F. \) How much longer does the turkey need to cook?
