Radioactivity was discovered in 1896 – so long ago that it now seems rather ordinary. Actually, it’s a remarkable thing; the nucleus of the atoms of some elements spontaneously explodes. This is a random event for the atom, and has nothing to do with how long the atom has existed, or with anything external to the atom. It just happens, although the disintegration rate is different for each type of radioactive atom. In fact, the rate is different for each isotope of an element; the number of neutrons in the nucleus makes a difference in how stable it is.
The shrapnel that emerges includes an alpha particle (2 protons and 2 neutrons, which is a helium nucleus) traveling at about 1/10 the speed of light. Alpha particles have enough energy that they can be observed individually as little flashes of light, emitted when the particle hits a suitable target.
After the disintegration, the nucleus has a smaller atomic number than the original, so that the atom has changed to another element (which may itself be radioactive).
In this post, I will just review some of the basic math involved with radioactivity. Radioactivity is a perfect example of exponential decay, which is an important thing for a mathematically literate person to know about.

Half Life

The most common way to quantify the disintegration rate is by stating a half life: the length of time, call it \(T\), we must wait before half the atoms in a sample have decayed. If \(N_0\) is the number of atoms we start with, then the number left after a time \(t\) is
\[N=N_0 \left( \frac{1}{2}\right)^{t/T}=N_0 \: 2^{-t/T} \tag{1}\]
This formula is intuitively sensible. When \(t=T\), we have \(N=N_0 (1/2)^1 = N_0 / 2 \), as required. With \(t=2T\), we get \(N=N_0 /4 \), etc.
It is more common and useful to write (1) as an exponential function with a base of e, instead of 2. This is easily done. Recall that \(\ln{2}\) is, by definition, a number such that e raised to that power equals 2. That is,
\[2=e^{\ln{2}}\]
We can substitute this for 2 in equation (1), and we get:
\[N=N_0 \left[e^{\ln{2}}\right]^{-t/T}=N_0 e^{(-\ln{2}/T) t} \tag{2}\]
We immediately see how to find the coefficient of the time value in the exponent; it’s just \(\ln{2}/T\).
Here we have changed the base of the exponential function from 2 to \(e\). Using similar methods, we can make the base of an exponential function anything we please by just rescaling the exponent.

The Age of the Earth

Until about 100 years ago, very little was known about the age of the Earth, with estimates ranging from a few million years to a few hundred million. In 1906, Ernest Rutherford suggested that radioactive decay might help to determine the Earth’s age by dating its rocks. One of his ideas was to determine how much helium was present in a rock that contained uranium. Since alpha particles are helium nuclei, the number of helium atoms in the rock should equal the number of uranium atoms that have decayed since the rock was formed. By comparing this with the number of uranium atoms left, he hoped to determine the length of time since the rock formed.
That particular idea did not work out well, but other radioactive-based methods have become almost the sole means we have for dating the Earth and the solar system. As one example of what can be done, consider the decay of uranium. Natural uranium consists of two isotopes:

• U-238, with 92 protons and 146 neutrons. U-238 has a half-life of \(4.468 \times 10^9\) years.
• U-235, with 92 protons and 143 neutrons. U-235 has a half-life of \(7.1 \times 10^8\) years, and it makes up less than one percent of Earth’s uranium. The actual ratio of the two is \(U_{235} / U_{238} = 0.0072\).

Like all elements heavier than iron, uranium is formed inside a supernova explosion. Physicists believe that this primeval uranium consists of the two isotopes in near equal portions. The Earth’s uranium has a small fraction of U-235 today only because because it decays more quickly than U-238.
We can easily find out how long it would take for the isotope proportions to change from equal parts to their presents levels. If we started with a unit quantity of each isotope, then the amount of each as a function of time t is:
\[\begin{align*} U_{235}=e^{\left( - \ln{2} / T_{235}\right) t} \\ U_{238}=e^{\left( - \ln{2} / T_{238}\right) t} \end{align*} \]
We know the present ratio of these two, so we can set:
\[e^{\left( - \ln{2} / T_{235}\right) t} / e^{\left( - \ln{2} / T_{238}\right) t} = 0.0072 \tag{3}\]
We can solve this for the unknown value of \(t\), (details in Note 1), and the result is \(t=6.0 \times 10^9\), or 6.0 billion years. This is a sensible value, because we know from other radioactivity methods that the solar system is 4.5 billion years old. The solar system apparently formed, at least in part, of matter from a supernova that occurred 1.5 billion years earlier.

Determining the Half Life

The example of uranium leads to a question: how can we even determine its half life? If the half life is in the neighborhood of a billion years, the tiny fraction that decays in a year, or even one hundred years, would be unmeasurable. This means that we cannot directly determine how much of a sample disintegrates during a reasonable time.
The solution is that we can actually count the number of atoms that decay during some time period, and from this we can compute the half life. We can count these atoms because each disintegration emits an alpha particle, and a detector near the sample will register each particle.
Here is how it works in detail: first we calculate the number of atoms in the sample. If the sample is U-238, we know that 238 grams of the material (e.g., one mole) has Avogadro’s number of atoms, where Avogadro’s number is \(A=6.023 \times 10^{23}\). Thus, if the sample is 1 gram, it contains \(N_0=A / 235\) atoms.
Now, suppose we count the number of alpha particles, call it \(n\), during some time \(\Delta t\). In terms of the equation for half life, that count, the number of atoms that disintegrate, would be
\[N_0-N_0 e^{(-\ln{2}/T)\Delta t} = n\]
Everything in this equation is known except T, the half life, so that we can solve for it.

Note 1

We can solve equation (3) for the time \(t\) as follows:
\[e^{\left( - \ln{2} / T_{235}\right) t} / e^{\left( - \ln{2} / T_{238}\right) t} = 0.0072 = e^{ - \ln{2}\left(1 / T_{235} - 1 / T_{238}\right) t}\]
As usual when the unknown is in an exponent, we take the logarithm of both sides:
\[ - \ln{2}\left(1 / T_{235} - 1 / T_{238}\right) t = \ln{0.0072} \]
or
\[t = \frac{\ln{0.0072}}{ - \ln{2}\left(1 / T_{235} - 1 / T_{238}\right)}\]
Using the above values for the half-lives, we get \(t=6.0 \times 10^9\) years.

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This entry was posted on Saturday, June 2nd, 2012 at 11:25 pm and is filed under Physics, Trig/PreCalculus. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.