Question

After a tree has been chopped down and burned to ash, the carbon isotopes in the ash are found to have a \({ }_{6}^{14} \mathrm{C}\) to \({ }_{6}^{12} \mathrm{C}\) ratio of \(1.300 \cdot 10^{-12} .\) Experimental tests on the \({ }_{6}^{14} \mathrm{C}\) atoms reveal that \({ }_{6}^{14} \mathrm{C}\) is a beta emitter with a half-life of 5730 yr. At an archeological excavation, a skeleton is found next to some wood ash from a campfire. If \(50.0 \mathrm{~g}\) of carbon from the ash emits electrons at a rate of 20.0 per hour, how long ago did the campfire burn?

Short answer

Answer: To find the years elapsed since the campfire burned, first find the initial amount of \({}_{6}^{14}\mathrm{C}\), the decay constant, the rate of decay in terms of activity, and the present number amount of \({}_{6}^{14}\mathrm{C}\) nuclei. Then, calculate the time using the formula \(t = -\frac{1}{\lambda} \cdot \ln{\frac{N}{N_{0}}}\).

Step-by-step solution

Determine the initial amount of \({}_{6}^{14}\mathrm{C}\)

We know that the initial ratio of \({}_{6}^{14}\mathrm{C}\) to \({}_{6}^{12}\mathrm{C}\) is \(1.300 \cdot 10^{-12}\). The total amount of carbon present is 50g. We can use this information to find the initial amount of \({}_{6}^{14}\mathrm{C}\) as follows: Let \(N_{0}\) be the initial amount of \({}_{6}^{14}\mathrm{C}\) in the wood ash. Then we have: \(N_{0} = 50 \times 1.300 \cdot 10^{-12}\)

Determine the decay constant

The half-life of \({}_{6}^{14}\mathrm{C}\) is 5730 years. We can use this information to find the decay constant (λ) as follows: \(\lambda = \frac{\ln{2}}{t_{\frac{1}{2}}}\), where \(t_{\frac{1}{2}}\) is the half-life of \({}_{6}^{14}\mathrm{C}\).

Determine the rate of decay in terms of activity (A)

We are given that the rate of electron emission is 20 per hour. We can use this information to determine the rate of decay (A) of \({}_{6}^{14}\mathrm{C}\) in the wood ash by multiplying the rate of electron emission by the Avogadro constant (N_A): \(A = 20 \frac{disintegration \ per \ hour \ per \ gram}{50.0 \ gram} = \frac{20}{50}\) disintegrations per hour

Determine the present number of \({}_{6}^{14}\mathrm{C}\) nuclei (N)

Now we will determine the present number of \({}_{6}^{14}\mathrm{C}\) nuclei (N) in the wood ash using the decay constant (λ) and the rate of decay (A): \(A = \lambda \cdot N\)

Calculate the time that has passed since the campfire burned (t)

Finally, we will determine the time that has passed since the campfire burned by using the initial number of \({}_{6}^{14}\mathrm{C}\) nuclei (N_0) and the present number of \({}_{6}^{14}\mathrm{C}\) nuclei (N) and the decay constant (λ) as follows: \(t = -\frac{1}{\lambda} \cdot \ln{\frac{N}{N_{0}}}\) By plugging in the values found in the previous steps, we can find the years since the campfire burned.

Key concepts

Carbon Isotopes

Carbon isotopes are variants of the carbon element that have the same number of protons but a different number of neutrons. The most common isotope, carbon-12 (₆¹²C), has six protons and six neutrons. Another stable isotope is carbon-13 (₆¹³C), with six protons and seven neutrons.

However, carbon-14 (₆¹⁴C) is different because it is radioactive. It has six protons and eight neutrons, making it unstable and prone to radioactive decay over time. In living organisms, carbon-14 is generated in the upper atmosphere and assimilated through the carbon dioxide cycle. Once the organism dies, it no longer absorbs carbon-14, and the isotope begins to decay at a predictable rate, which is used in carbon dating.

The exercise presents a scenario where the carbon isotopes in wood ash are analyzed for carbon dating purposes. We monitor the ratio of ₆¹⁴C to ₆¹²C to estimate the time since the wood was burned.

Radioactive Decay

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. This spontaneous transformation can result in a change of the element, as the original atom alters in composition. There are many types of radioactive decay; one common type is beta decay, where a neutron converts into a proton while releasing an electron (a beta particle) and an antineutrino.

In the given exercise, carbon-14 exhibits beta decay, slowly reducing its quantity over time. The rate at which this process occurs is indicated by the decay constant, represented by the Greek letter lambda (\( \lambda \)). The decay constant is a key parameter in calculating the age of archaeological samples, as it factors into determining the amount of time passed since the death of a biological sample, such as the burning of the wood in our exercise.

Half-life

The half-life of a radioactive isotope is the time required for half of the isotope in a sample to decay. This is a constant value for each radioactive substance and is unaffected by environmental factors like temperature or pressure. In simpler terms, after one half-life, you will have half of the original radioactive atoms remaining, after two half-lives, one quarter will remain, and so on.

The half-life of carbon-14, as mentioned in the exercise, is 5730 years. With this constant, scientists can determine how many years have passed since the isotope ceased to be replenished—effectively the time since the organism or wood stopped living. By comparing the current activity (rate of decay) to the initial activity assumed to be present in the living sample, we can apply the concept of half-life to determine the age of the archaeological find, which is the ultimate aim of the exercise.