Applying the Integrated Rate Law for Radioactive Decay: Radiocarbon Dating A wooden object found in an Indian burial mound is subjected to radiocarbon dating. The activity associated with its ^14C content is 10 dis min^-1 g^-1. What is the age of the object? In other words, how much time has

Question

Applying the Integrated Rate Law for Radioactive Decay: Radiocarbon Dating

A wooden object found in an Indian burial mound is subjected to radiocarbon dating. The activity associated with its [latex]_{}^{14} ext{C}[/latex] content is 10 dis [latex] ext{min}^{-1} ext{ g}^{-1}[/latex]. What is the age of the object? In other words, how much time has elapsed since the tree from which the wood came was cut down?

Analyze
The solution requires three equations: (25.11), (25.12), and (25.13).

rate of decay ∝ N and rate of decay = A = λN (25.11)

[latex]\ln \left(\frac{N_t}{N_0} ight)=-\lambda t[/latex] (25.12)

[latex]t_{1 / 2}=\frac{0.693}{\lambda}[/latex] (25.13)

Accepted Answer

Equation (25.13) is used to determine the decay constant.

[latex]\lambda=\frac{0.693}{5730 y}=1.21 imes 10^{-4} y ^{-1}[/latex]

Next, equation (25.11) relates to the actual number of atoms: N at t = 0 (the time when the [latex]_{}^{14} ext{C}[/latex] equilibrium was destroyed) and [latex]N_t[/latex] at time t (the present time). As discussed on page 1123, the activity just before the [latex]_{}^{14} ext{C}[/latex] equilibrium was destroyed was 15 dis [latex] ext{min}^{-1} ext{ g}^{-1}[/latex]; at the time of the measurement, it is 10 dis [latex] ext{min}^{-1} ext{ g}^{-1}[/latex]. The corresponding numbers of atoms are equal to these activities divided by λ.

[latex]N_0=A_0 / \lambda=15 / \lambda \quad ext { and } \quad N_t=A_t / \lambda=10 / \lambda[/latex]

Finally, we substitute into equation (25.12).

[latex]\ln \frac{N_t}{N_0}=\ln \frac{10 / \lambda}{15 / \lambda}=\ln \frac{10}{15}=-\left(1.21 imes 10^{-4} y ^{-1} ight) t[/latex]

[latex]-0.41=-\left(1.21 imes 10^{-4} y^{-1} ight) t[/latex]

[latex]t=\frac{0.41}{1.21 imes 10^{-4} y ^{-1}}=3.4 imes 10^3 y[/latex]

Assess
In the previous example, we observed that the activity depends on the amount of material. Therefore, the results of radiocarbon dating depend on knowing the activity at the time the equilibrium between [latex]_{}^{14} ext{C}[/latex] and the other nonradioactive carbon isotopes ceases. If at the time equilibrium was destroyed the activity was 14 dis [latex] ext{min}^{-1} ext{ g}^{-1}[/latex], we would have determined the object to be 2.8 × 10³ years old, which is approximately a 17% error.

Question 25.4: Applying the Integrated Rate Law for Radioactive Decay: Radi...

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