Question

A rock from an archaeological dig was found to contain \(0.255 \mathrm{~g}\) of Pb-206 per gram of U-238. Assume that the rock did not contain any Pb-206 at the time of its formation and that U-238 decayed only to Pb-206. How old is the rock? (For \(\mathrm{U}-238, t_{1 / 2}=4.5 \times 10^{9} \mathrm{y}\).)

Short answer

Question: Based on the given proportion of Pb-206 to U-238 in the rock, determine the age of the rock. Answer: The rock is approximately \(2.53 \times 10^9\) years old.

Step-by-step solution

Define the half-life formula

For radioactive decay, we can define the half-life formula as follows: \(N(t) = N_0 e^{-\lambda t}\), where - \(N(t)\) is the number of undecayed nuclei at time \(t\) - \(N_0\) is the initial number of undecayed nuclei - \(\lambda\) is the decay constant (\(\lambda = \frac{\ln{2}}{t_{1/2}}\), where \(t_{1/2}\) is the half-life) - \(t\) is the time elapsed

Calculate the decay constant

Given the half-life of U-238, we can calculate the decay constant, \(\lambda\), using the formula: \(\lambda = \frac{\ln{2}}{t_{1/2}}\) Plugging in the given half-life for U-238 (\(t_{1/2}=4.5 \times 10^{9}\mathrm{y}\)), we get: \(\lambda = \frac{\ln{2}}{4.5 \times 10^{9}\mathrm{y}} = 1.54 \times 10^{-10} \mathrm{y}^{-1}\)

Set up the equation to find time

Using the given proportion of \(\frac{0.255\,\mathrm{g\,Pb-206}}{1\,\mathrm{g\,U-238}}\), we can set up an equation to solve for the time \(t\). Since there was no initial Pb-206 in the rock, we can write the proportion as: \(\frac{N_0 - N(t)}{N(t)} = \frac{0.255}{1}\) Here \(N_0 - N(t)\) is the amount of U-238 that has decayed to Pb-206.

Solve for the remaining U-238

Now, let's rearrange the equation to find \(N(t)\), the remaining U-238: \(N(t) = \frac{N_0}{1 + 0.255} = \frac{N_0}{1.255}\)

Find the elapsed time t

Now, use the half-life formula to solve for \(t\): \(N(t) = N_0 e^{-\lambda t}\) \(\frac{N_0}{1.255} = N_0 e^{-1.54 \times 10^{-10}t}\) Divide both sides by \(N_0\): \(\frac{1}{1.255} = e^{-1.54 \times 10^{-10}t}\) Take the natural logarithm of both sides: \(-\ln{1.255} = -1.54 \times 10^{-10}t\) Solve for \(t\): \(t = \frac{\ln{1.255}}{1.54 \times 10^{-10}} \approx 2.53 \times 10^9 \mathrm{y}\) The rock is approximately \(2.53 \times 10^9\) years old.

Key concepts

Understanding the Half-Life Formula

The concept of half-life is central to the field of radioactive dating. It describes the time required for half of the radioactive nuclei in a sample to decay. The half-life formula is essential for calculating the age of archaeological samples and understanding earth’s geological events. The formula for the half-life is often written in the form \[\begin{equation}N(t) = N_0 e^{-lambda t},\text{where}\begin{itemize}\item \(N(t)\) represents the number of undecayed nuclei at time \(t\),\item \(N_0\) is the initial count of undecayed nuclei,\item \(\lambda\) denotes the decay constant, and\item \(t\) stands for the elapsed time.\text{​}\text{​}<\/itemize>\end{equation}\]To interpret this correctly, it’s important to recognize that as time progresses, the count of undecayed nuclei dwindles as they transform into other elements. The half-life formula shows that the relationship between the remaining undecayed nuclei and time is exponential in nature. In practice, understanding this relationship allows us to backtrack and estimate the age of a sample by comparing the remaining undecayed nuclei to the initial amount and accounting for the decay constant (lambda).

When students engage with half-life problems, visual aids such as decay curves can provide a clearer picture of the exponential decay process. It's also crucial for learners to be comfortable with the natural logarithm and exponential functions, as these are integral to manipulating the half-life formula.

The Role of Decay Constant in Radioactive Dating

The decay constant (\lambda) plays a pivotal role in the field of geochronology and the decay process of radioactive isotopes. It quantifies the probability of decay of a nucleus per unit time. In essence, a higher decay constant indicates a faster decay rate. The decay constant is intimately linked to the half-life (), with the mathematical relationship being \[\begin{equation}lambda = \frac{\ln{2}}{t_{1/2}}.\end{equation}\]This equation states that the decay constant is the natural logarithm of 2 (approximately 0.6931) divided by the half-life of the isotope. The natural logarithm emerges from the exponential decay formula, representing a continuous decay process.

For students, it’s important to grasp how \lambda affects radioactive dating calculations. Because \lambda is derived from the known half-life, it remains a constant value throughout the decay process. Its consistent nature provides the reliability required to calculate dates using the half-life formula. During exercises, students can improve comprehension by calculating decay constants with different half-lives to observe how they influence the rate of decay.

Radiometric Age Calculation

Radiometric age calculation is the process of determining the age of a material based on the amount of a radioactive isotope and its decay product present in the sample. By combining the concepts of the half-life and decay constant, we can solve the equation to find the elapsed time since the isotope began to decay, which is indicative of the sample's age. This method is widely used in geology, archaeology, and anthropology to date rocks and artifacts.

To calculate the age, derive an equation starting with the proportion of the parent and daughter isotopes. By expressing this ratio in terms of the half-life formula, rearranging and solving for the time (), we can infer when the sample was formed. A typical equation used for this purpose is:
\[\begin{equation}N(t) = N_0 e^{-\lambda t},\end{equation}\]By rearranging and isolating , the elapsed time -- or the age of our sample -- can be found. As an example, in radiometric dating exercises, students might solve for the time , based on the known decay product ratio. It's essential for learners to carry out the calculation steps methodically, paying close attention to the units and making sure that exponential and logarithmic operations are performed correctly. Having sample problems with varying degrees of complexity can aid in deepening the understanding of the radiometric age calculation.