Question

A rock contains 0.688 \(\mathrm{mg}^{206} \mathrm{Pb}\) for every 1.000 \(\mathrm{mg}^{238} \mathrm{U}\) present. Assuming that no lead was originally present, that all the \(^{206}\mathrm{P}\) formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between \(^{238} \mathrm{U}\) and \(^{206 \mathrm{P}} \mathrm{b}\) is negligible, calculate the age of the rock. (For \(38 \mathrm{U}, t_{1 / 2}=4.5 \times 10^{9}\) years.)

Short answer

The age of the rock is approximately \(3.78 \times 10^9\) years.

Step-by-step solution

Understand the decay formula and variables

The decay formula is given by: \[ N_t = N_0 e^{-\lambda t} \] Where: - \(N_t\) is the number of nuclides remaining at time \(t\). - \(N_0\) is the initial number of nuclides. - \(\lambda\) is the decay constant. - \(t\) is time which we want to find in this problem. After understanding the decay formula and its variables, let's proceed to find the decay constant.

Calculate the decay constant (\(\lambda\))

The decay constant is related to the half-life as follows: \[ \lambda = \frac {\ln 2} {t_{1/2}} \] The half-life of \(^{238}\mathrm{U}\) is given as \(t_{1/2}=4.5 \times 10^9 \ \mathrm{years}\). Let's find the decay constant: \[ \lambda = \frac {\ln 2} {4.5 \times 10^9 \ \mathrm{years}} \approx 1.54 \times 10^{-10} \ \mathrm{year}^{-1}\] Now we have the decay constant of \(^{238}\mathrm{U}\). Let's move to the next step.

Derive the equation relating the initial and remaining nuclides of \(^{238}\mathrm{U}\) and \(^{206}\mathrm{Pb}\)

We know that: \[ N_t = N_0 e^{-\lambda t} \] Let \(N_U\) and \(N_{Pb}\) be the remaining amounts of \(^{238}\mathrm{U}\) and \(^{206}\mathrm{Pb}\) respectively. Initially, there was no \(^{206}\mathrm{Pb}\) and the amount of \(^{238}\mathrm{U}\) is \(N_0\). So, we can write: \[ N_{Pb} = N_0 - N_U \] As given, the ratio of \(^{206}\mathrm{Pb}\) to \(^{238}\mathrm{U}\) in the rock is \(\frac{0.688}{1}\). Thus, \[ \frac{N_{Pb}}{N_U} = 0.688 \] We can substitute \(N_{Pb}\) from the first equation: \[ \frac {N_0 - N_U}{N_U} = 0.688 \] Now that we have an equation for the initial and remaining \(^{238}\mathrm{U}\) and \(^{206}\mathrm{Pb}\), let's solve for \(N_U\).

Calculate the remaining amount of \(^{238}\mathrm{U}\) (\(N_U\))

Let's solve the equation we derived in step 3 for \(N_U\): \[ N_0 - N_U = 0.688 N_U \] \[ N_0 = 1.688 N_U \] \[ N_U = \frac {N_0}{1.688} \] Let's substitute this value in the decay formula for \(N_t\): \[ N_t = N_0 e^{-\lambda t} \] \[ \frac {N_0}{1.688} = N_0 e^{-\lambda t} \] Now, we can solve for time \(t\).

Calculate the age of the rock (time \(t\))

We have the decay formula: \[ \frac {N_0}{1.688} = N_0 e^{-\lambda t} \] Divide both sides by \(N_0\): \[ \frac {1}{1.688} = e^{-\lambda t} \] Take the natural logarithm of both sides: \[ \ln \frac {1}{1.688} = -\lambda t \] Now, solve for \(t\): \[ t = \frac {\ln \frac {1}{1.688}}{-\lambda} \] Substitute the value of \(\lambda\): \[ t = \frac {\ln \frac {1}{1.688}}{-1.54 \times 10^{-10} \ \mathrm{year}^{-1}} \approx 3.78 \times 10^9 \ \mathrm{years} \] So, the age of the rock is approximately \(3.78 \times 10^9\) years.

Key concepts

Half-life Calculations

To understand how old a rock is by looking at uranium-lead dating, we first need to grasp the concept of half-life. Half-life is the time it takes for half of a given amount of a radioactive substance to decay. For example, if you start with 1 gram of uranium, after one half-life, only 0.5 grams of uranium will remain undecayed. The decay process continues exponentially, which means after two half-lives, you'd have 0.25 grams of uranium left.
Notably, in the case of Uranium-238 — the isotope used in our problem — its half-life is 4.5 billion years (4.5 imes 10^9 ext{ years}). This lengthy period makes it an excellent choice for dating ancient rocks.
Calculating the half-life involves using the natural exponential function: N_t = N_0 e^{-rac{ ext{ln} 2}{t_{1/2}} t}, where \(N_t\) represents the remaining amount of the isotope, \(N_0\) represents the initial amount, and \(t\) is the time that has passed. This relationship provides a foundational understanding necessary for determining how old certain rocks are based on uranium content.

Decay Constant

The decay constant, often represented by the Greek letter \(\lambda\), is a fundamental concept when discussing radioactive decay. It quantifies the rate at which a radioactive isotope decays. In mathematical terms, it relates to the half-life through the formula \(\lambda = \frac{\ln 2}{t_{1/2}}\). This formula allows us to determine the decay constant once the half-life is known and vice versa.
In the context of Uranium-238, for instance, with its half-life of \(4.5 \times 10^9\) years, we find its decay constant to be around \(1.54 \times 10^{-10} \, \text{year}^{-1}\). This constant is crucial because it lets us calculate how a substance decreases over time, which is vital for applications like uranium-lead dating.
The decay constant is used in the exponential decay formula \(N_t = N_0 e^{-\lambda t}\), indicating how the quantity of a radioactive element diminishes over time. Understanding this concept enables us to make predictions about the age of samples such as rocks or archaeological finds.

Uranium-Lead Dating

Uranium-lead dating is a robust method for determining the age of geological formations. It hinges on the radioactive decay of Uranium-238 into Lead-206. This transformation occurs because of the decay steps that Uranium-238 undergoes, eventually forming stable Lead-206. The method is favored for rock dating because of the long half-life of Uranium-238 — \(4.5 \times 10^9\) years.
When a rock forms, it contains uranium but no lead. Over time, uranium decays into lead. By measuring the ratio of lead to uranium in the rock, scientists can calculate how long this conversion has been occurring, thus estimating the rock's age. This is done using the decay equation \(N_t = N_0 e^{-\lambda t}\).
In practical terms, if we have a rock with a ratio of 0.688 mg of Lead-206 for every 1 mg of Uranium-238, we can use the decay constant derived from its half-life to calculate the age. Applying the natural logarithm and the decay formula helps us solve for the time \(t\), providing an estimated age of approximately \(3.78 \times 10^9\) years for the rock in question.