## Content

### Radioactive decay and half-life

#### Decay of carbon-14

Carbon-14 is a radioactive isotope of carbon, containing 6 protons and 8 neutrons, that is present in the earth's atmosphere in extremely low concentrations.^{2}

It is naturally produced in the atmosphere by cosmic rays (and also artificially by nuclear weapons), and continually decays via nuclear processes into stable nitrogen atoms.

Suppose we have a sample of a substance containing some carbon-14. Let \(m\) be the mass of carbon-14 in nanograms after \(t\) years. ^{3}

It turns out that, if the sample is isolated, then \(m\) and \(t\) approximately ^{4} satisfy the differential equation

Suppose our sample initially contains 100 nanograms of carbon-14. Let's investigate what happens to the sample over time.

First, we can solve the differential equation. Since \(m\) has a continuous decay rate of \(-0.000121\), a general solution to the differential equation is

\[ m(t) = C e^{-0.000121 \, t}, \]where \(C\) is a constant. Substituting the initial condition \(t = 0\), \(m = 100\) gives \(C = 100\), so

\[ m(t) = 100 \, e^{-0.000121 \, t}. \]With this formula, we can calculate the amount \(m\) of carbon-14 over the years.

\(t\) (years) | \(m\) (ng to 4 decimal places) |
---|---|

0 | 100.0000 |

100 | 98.7973 |

1000 | 88.6034 |

2000 | 78.5056 |

5000 | 54.6074 |

\(10\ 000\) | 29.8197 |

\(20\ 000\) | 8.8922 |

Every year, the mass \(m\) of carbon-14 is multiplied by \(e^{-0.000121} \approx 0.999879\). After 100 years, 98.7973 nanograms still remain. After 1000 years, we still have 88.6034 nanograms. But after 5000 years, however, almost half of the carbon-14 has decayed.

#### Half-life of carbon-14

#### Example

How long does it take for precisely half of the carbon-14 in the sample to decay; that is, when does \(m=50\)? Give the answer to three significant figures.

#### Solution

The mass of carbon-14 in our sample is given by

\[ m(t) = 100 \, e^{-0.000121 \, t}. \]So we solve \(50 = 100 \, e^{-0.000121 \, t}\), which gives \(e^{-0.000121 \, t} = \dfrac{1}{2}\). Hence,

\[ t = \dfrac{\log_e \dfrac{1}{2}}{-0.000121} \approx 5730 \text{ years (to three significant figures).} \]The time period calculated in this example is called the **half-life** of carbon-14. In fact over any period of 5730 years, the amount of carbon-14 in an isolated sample will decay by half. This fact is used in **radiocarbon dating** to determine the age of fossils up to \(60\ 000\) years old. Roughly speaking, while an organism is alive, its interactions with its environment maintain a constant ratio of carbon-14 to carbon-12 in the organism; but after it dies, the carbon-14 is no longer replenished, and the ratio of carbon-14 to carbon-12 decays in a predictable way. (See Wikipedia for more on radiocarbon dating.)

Exercise 1

Explain why the mass of carbon-14 in the sample is given (approximately) by

\[ m(t) = 100 \, \Bigl( \dfrac{1}{2} \Bigr)^{\tfrac{t}{5730}}, \]and hence explain why the amount of carbon-14 in the sample decays by half over any period of 5730 years.

#### Half-life in general

In general, whenever a quantity \(x(t)\) obeys an exponential decay equation

\[ x(t) = C e^{kt}, \]where the continuous decay rate \(k\) is negative, then the quantity \(x\) has a half-life \(T\). After any time period of length \(T\), the quantity \(x\) decreases by half. Let us see why.

As \(k\) is negative, the factor \(e^{kt}\) decreases from 1 (at \(t=0\)) towards 0 (as \(t\) approaches \(\infty\)). Therefore there is a time \(t=T\) such that

\[ e^{kT} = \dfrac{1}{2}. \]We now solve for \(T\) and obtain

\begin{align*} kT &= \log_e \dfrac{1}{2} \\ &= - \log_e 2, \end{align*} so \[ T = - \dfrac{1}{k} \log_e 2. \]This \(T\) is the half-life. From time \(t=0\) to time \(t=T\), the factor \(e^{kt}\) decreases from \(e^0 = 1\) to \(e^{kT} = \dfrac{1}{2}\), that is, decreases by half. Similarly, over any time period of length \(T\), the term \(e^{kt}\) decreases by half. ^{5}

Note that, when \(k = -0.000121\), we obtain \(T = 5730\), in agreement with our calculation for carbon-14.