Problem 29 - Calculus Early Transcendentals

Question

Devise an exponential decay function that fits the following data; then answer the accompanying questions. Be sure to identify the reference point \((t=0)\) and units of time. Uranium-238 (U-238) has a half-life of 4.5 billion years. Geologists find a rock containing a mixture of \(\mathrm{U}-238\) and lead, and determine that \(85 \%\) of the original U-238 remains; the other \(15 \%\) has decayed into lead. How old is the rock?

Step-by-Step Solution

Answer

Answer: The rock is approximately 1.05 billion years old.

Step 1: Devise the exponential decay function

The general formula for an exponential decay function is of the form \(N(t)=N_0 e^{-\lambda t}\), where \(N(t)\) is the amount of the substance remaining after time \(t\), \(N_0\) is the initial amount of the substance, \(\lambda\) is the decay constant, and \(t\) is time. In the case of Uranium-238, we know that the half-life is 4.5 billion years. We can determine the decay constant using the formula for half-life: \(T_{1/2} = \frac{\ln{2}}{\lambda}\).

Step 2: Find the decay constant

Using the given half-life of U-238 (4.5 billion years), we can find the decay constant. Re-arrange the half-life formula to solve for \(\lambda\): \(\lambda = \frac{\ln{2}}{T_{1/2}}\) Substituting the value for half-life: \(\lambda = \frac{\ln{2}}{4.5\,\text{billion years}} \approx 1.54 \times 10^{-10}\, \text{year}^{-1}\)

Step 3: Define the exponential decay function for U-238

Now that we have the decay constant, we can define the exponential decay function for U-238: \(N(t) = N_0 e^{-1.54 \times 10^{-10}t}\) Here, \(N_0\) represents the initial amount of U-238 and \(t\) represents the time in years. The reference point \(t=0\) corresponds to when the rock was formed.

Step 4: Determine the age of the rock using the decay function

The geologists found that 85% of the original U-238 remains. We can write this as a ratio, so \(N(t) = 0.85N_0\). We want to find the time \(t\) when this occurs. Plugging this ratio into the decay function and solving for \(t\): \(0.85N_0 = N_0 e^{-1.54 \times 10^{-10}t}\) Divide both sides by \(N_0\): \(0.85 = e^{-1.54 \times 10^{-10}t}\) Now, take the natural logarithm of both sides to isolate \(t\): \(\ln{0.85} = -1.54 \times 10^{-10}t\) Solve for \(t\): \(t = \frac{\ln{0.85}}{-1.54 \times 10^{-10}} \approx 1.05\,\text{billion years}\) The rock is approximately \(1.05\) billion years old.

Key Concepts

Uranium-238

Uranium-238, often abbreviated as U-238, is a naturally occurring isotope of uranium. It is the most abundant uranium isotope, constituting about 99.3% of natural uranium.
U-238 is not used directly as a fuel in nuclear reactors, unlike Uranium-235, which is fissile. However, U-238 plays a crucial role in nuclear technology as it can be converted into Plutonium-239, which is a fissile isotope, through the process of neutron capture and subsequent beta decay.

• Abundance: 99.3% of natural uranium.
• Conversion: Can be converted into Plutonium-239, a useful nuclear fuel.
• Importance: Key component in nuclear power and weapons technology.

Understanding Uranium-238's properties is essential for geologists, as its radioactive decay is used to date the age of rocks and geological formations.

half-life

The concept of half-life is fundamental in the study of radioactive decay. The half-life of a radioactive isotope is the time required for half of the initial quantity of the isotope to decay. It is a measure of the rate at which the substance undergoes radioactive decay.
For Uranium-238, the half-life is remarkably long, at 4.5 billion years. This means it takes 4.5 billion years for half of a given sample of U-238 to decay into lead.

• Definition: Time taken for half of the isotope to decay.
• U-238 Half-life: 4.5 billion years.
• Usage: Used in radiometric dating methods to determine the age of rocks and fossils.

The lengthy half-life makes U-238 an ideal candidate for dating geological formations that are billions of years old, providing insights into Earth's history.

decay constant

The decay constant, often denoted by the symbol \(\lambda\), is a crucial parameter in the exponential decay model of radioactive substances. It represents the probability per unit time that a nucleus will decay.
For instance, if Uranium-238 has a half-life of 4.5 billion years, its decay constant \(\lambda\) can be calculated using the relationship:
\[\lambda = \frac{\ln{2}}{T_{1/2}}\]
For U-238, this evaluates to approximately \(1.54 \times 10^{-10}\, \text{year}^{-1}\). The decay constant is vital in determining how quickly a substance will decay over time.

• Symbol: \(\lambda\).
• Formula: \(\lambda = \frac{\ln{2}}{T_{1/2}}\).
• U-238 Decay Constant: Approximately \(1.54 \times 10^{-10}\, \text{year}^{-1}\).

Understanding the decay constant allows us to model and predict the behavior of radioactive substances accurately, which is essential in fields like geology and nuclear physics.

natural logarithm

The natural logarithm is a fundamental mathematical function that arises frequently in exponential growth and decay calculations. It is denoted by \(\ln\) and is the logarithm to the base \(e\), where \(e\) is approximately 2.71828.
In the context of radioactive decay, the natural logarithm plays a crucial role in solving for the time \(t\) when you have specific decay information. For instance, if you know the remaining fraction of U-238 in a sample, you can use the natural logarithm to solve for the time passed:
\[\ln{\left(\frac{N(t)}{N_0}\right)} = -\lambda t\]
This equation allows you to calculate the age of a rock sample when a certain percentage of Uranium-238 remains.

• Symbol: \(\ln\).
• Base: \(e\), approximately 2.71828.
• Usage: Key in calculating decay rate and time in exponential decay problems.

The natural logarithm is an essential tool for many scientific calculations, especially in fields dealing with exponential relationships like physics and chemistry.