Problem 31 - Precalculus with Calculus Previews

Question

Potassium is one of the most abundant metals found throughout the Earth's crust and oceans. Although potassium occurs naturally in the form of three isotopes, only the isotope potassium- \(40(\mathrm{~K}-40)\) is radioactive. This isotope is unusual in that it decays by two different nuclear reactions. By emitting a beta particle a great percentage of an initial amount of K-40 decays over time into the stable isotope calcium-4o (Ca-4o), whereas by electron capture a smaller percentage of \(\mathrm{K}-40\) decays into the stable isotope argon- \(40(\mathrm{Ar}-40)\). If it is assumed that rates at which the amounts \(C(t)\) of \(\mathrm{Ca}-40\) and \(A(t)\) of Ar-40 increase are proportional to the amount \(P(t)\) of potassium present at time \(t\), and that the rate at which \(P(t)\) decays is also proportional to \(P(t),\) then it can be shown that \(C(t)=\frac{k_{C}}{k_{A}+k_{C}} P_{0}\left[1-e^{-\left(k_{A}+k_{C}\right) t}\right]\) \(A(t)=\frac{k_{A}}{k_{A}+k_{C}} P_{0}\left[1-e^{-\left(k_{A}+k_{C}\right) t}\right]\) \(P(t)=P_{0} e^{-\left(k_{A}+k_{C}\right) t}\) where \(P(0)=P_{0}, k_{C}=4.962 \times 10^{-10},\) and \(k_{A}=\) \(0.581 \times 10^{-10}\) (a) Find the half-life of \(\mathrm{K}-40\). (b) Determine the percentage of an initial amount \(P_{0}\) of \(\mathrm{K}-40\) that decays into \(\mathrm{Ca}-40\) and the percentage that decays into Ar-40 over a very long period of time, that is, as \(t \rightarrow \infty\).

Step-by-Step Solution

Answer

(a) The half-life of \( \text{K-40} \) is approximately \( 1.25 \times 10^9 \) years. (b) \( 89.54\% \) decays into \( \text{Ca-40} \) and \( 10.46\% \) into \( \text{Ar-40} \).

Step 1: Understanding Half-Life Equation

The half-life is the time it takes for a quantity to reduce to half its initial amount. The formula for half-life \( t_{1/2} \) is derived from the exponential decay function. Since we have \( P(t) = P_0 e^{-(k_C + k_A) t} \), let's set \( P(t) = \frac{P_0}{2} \). The goal is to solve for \( t \).

Step 2: Derive the Half-Life Formula

To find the half-life, set \( P(t) = \frac{P_0}{2} = P_0 e^{-(k_C + k_A) t_{1/2}} \). Divide both sides by \( P_0 \): \( \frac{1}{2} = e^{-(k_C + k_A) t_{1/2}} \). Take the natural logarithm on both sides: \( \ln\left(\frac{1}{2}\right) = -(k_C + k_A) t_{1/2} \). Solve for \( t_{1/2} \): \( t_{1/2} = \frac{\ln\left(2\right)}{k_C + k_A} \).

Step 3: Calculate the Half-Life of K-40

Substitute the given values of \( k_C = 4.962 \times 10^{-10} \) and \( k_A = 0.581 \times 10^{-10} \) into \( t_{1/2} = \frac{\ln(2)}{k_C + k_A} \). First, calculate \( k_C + k_A = 5.543 \times 10^{-10} \). Then, \( t_{1/2} = \frac{0.693}{5.543 \times 10^{-10}} \approx 1.25 \times 10^9 \) years.

Step 4: Determine Percentage of Decay Products

To find the percentage that decays into \( \text{Ca-40} \) and \( \text{Ar-40} \) over a very long time, refer to the limits \( t \to \infty \) of \( C(t) \) and \( A(t) \). From \( C(t) = \frac{k_C}{k_A+k_C} P_0 [1-e^{-(k_A+k_C) t}] \) and \( A(t) = \frac{k_A}{k_A+k_C} P_0 [1-e^{-(k_A+k_C) t}] \), observe that as \( t \to \infty \), \( 1-e^{-(k_A+k_C) t} \to 1 \).

Step 5: Calculate Percentages

The function \( C(t) \) simplifies to \( C(\infty) = \frac{k_C}{k_A+k_C} P_0 \). The percentage decay into \( \text{Ca-40} \) is \( \frac{k_C}{k_A+k_C} \times 100\% \). Similarly, \( A(\infty) = \frac{k_A}{k_A+k_C} P_0 \), hence \( \text{Ar-40} \)'s percentage is \( \frac{k_A}{k_A+k_C} \times 100\% \). Substitute the known values to compute: \( \frac{4.962}{5.543} \times 100\% \approx 89.54\% \) for \( \text{Ca-40} \) and \( \frac{0.581}{5.543} \times 100\% \approx 10.46\% \) for \( \text{Ar-40} \).

Key Concepts

Potassium-40

Potassium-40 (\(\text{K-40}\)) is a unique radioactive isotope of potassium, which is naturally found in small amounts on Earth. It is unlike the other isotopes because of its radioactivity, meaning it spontaneously transforms into different elements over time. What makes \(\text{K-40}\) particularly interesting is that it decays through two distinct types of nuclear reactions:

• Beta Decay: In this process, \(\text{K-40}\) emits a beta particle and transforms into calcium-40 (\(\text{Ca-40}\)), accounting for a significant portion of its decay.
• Electron Capture: Here, \(\text{K-40}\) captures an electron to become argon-40 (\(\text{Ar-40}\)), a process that occurs less frequently.

Like many radioactive isotopes, the decay of \(\text{K-40}\) is a naturally occurring process, contributing to dating geological events such as the age of rocks and fossils in the field of geochronology.

Half-Life Equation

The half-life, denoted as \(t_{1/2}\), is a crucial concept in understanding radioactive decay processes. It represents the time required for half of a sample of a radioactive isotope to decay, reducing it to half of its original quantity.For \(\text{K-40}\), the half-life equation is derived using the formula for exponential decay. Given that \( P(t) = P_0 e^{-(k_C + k_A) t} \) captures the decay of \(\text{K-40}\) over time, setting \(P(t) = \frac{P_0}{2}\) allows us to solve for \(t_{1/2}\).

• Start by dividing both sides by \(P_0\): \(\frac{1}{2} = e^{-(k_C + k_A) t_{1/2}}\).
• Take the natural logarithm of both sides to obtain \(\ln\left(\frac{1}{2}\right) = -(k_C + k_A) t_{1/2}\).
• Finally, solve for \(t_{1/2}\): \(t_{1/2} = \frac{\ln(2)}{k_C + k_A}\).

For \(\text{K-40}\), this results in a half-life of approximately 1.25 billion years, highlighting its invaluable role in dating ancient geological materials.

Exponential Decay Function

The exponential decay function is fundamental in modeling how radioactive isotopes like \(\text{K-40}\) decrease over time. It describes the decay rate based on the principle that the rate of decay is proportional to the quantity present.For \(\text{K-40}\), this exponential decay is represented by:\[P(t) = P_0 e^{-(k_C + k_A)t}\]Here:

• \(P(t)\) is the amount of \(\text{K-40}\) at time \(t\).
• \(P_0\) is the initial amount of \(\text{K-40}\).
• \(k_C\) and \(k_A\) are decay constants specific to the \(\text{Ca-40}\) and \(\text{Ar-40}\) decay processes, respectively.

As time progresses, the exponential term causes the quantity of \(\text{K-40}\) to approach zero. Additionally, the decay into \(\text{Ca-40}\) and \(\text{Ar-40}\) is also governed by similar exponential functions, each with a specific fraction of the decay constants:

• \(C(t) = \frac{k_C}{k_A + k_C} P_0 [1 - e^{-(k_A + k_C)t}]\) for \(\text{Ca-40}\).
• \(A(t) = \frac{k_A}{k_A + k_C} P_0 [1 - e^{-(k_A + k_C)t}]\) for \(\text{Ar-40}\).

These equations help identify the amount of each product formed over time, showing the complexity and beauty of radioactive decay.