Problem 19 - Applied Calculus

Question

One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at a rate of approximately \(2.5 \%\) per year. (a) Write the percent of strontium-90 remaining, \(P\), as a function of years, \(t\), since the nuclear accident. [Hint: \(100 \%\) of the contaminant remains at \(t=0 .]\) (b) Graph \(P\) against \(t\) (c) Estimate the half-life of strontium-90. (d) After the Chernobyl disaster, it was predicted that the region would not be safe for human habitation for 100 years. Estimate the percent of original strontium-90 remaining at this time.

Step-by-Step Solution

Answer

(a) \( P(t) = 100(0.975)^t \); (c) Half-life \(\approx 27.72\) years; (d) \(\approx 7.72\%\) remains after 100 years.

Step 1: Understanding the Exponential Decay Formula

The decay process can be modeled using the exponential decay formula \( P(t) = P_0 e^{-kt} \), where \( P(t) \) is the percentage of the substance remaining at time \( t \), \( P_0 \) is the initial amount, \( k \) is the decay constant, and \( e \) is the base of the natural logarithm.

Step 2: Determining the Decay Constant

The decay rate is given as 2.5% per year. To find \( k \), the decay constant, we use the formula \( k = \ln(1 - \, ext{decay rate}) \). Converting 2.5% to a decimal gives 0.025, so \( k = \ln(1 - 0.025) = \ln(0.975) \).

Step 3: Writing the Function for Strontium-90 Decay

Substitute \( P_0 = 100 \) (since 100% is the initial amount at \( t=0 \)) and the calculated \( k \) into the decay formula: \[ P(t) = 100e^{- ext{ln}(0.975)t} = 100(0.975)^t \].

Step 4: Graphing the Function

To graph \( P \) against \( t \), plot the exponential decay function \( P(t) = 100(0.975)^t \) for \( t \) ranging from 0 to 100 years. This will show a decreasing curve starting at 100 at \( t=0 \) and approaching zero as \( t \) increases.

Step 5: Estimating the Half-Life of Strontium-90

The half-life is the time it takes for half of the substance to decay. Set \( P(t) = 50 \) and solve \( 50 = 100(0.975)^t \). Simplify to \( 0.5 = (0.975)^t \). Taking the natural logarithm of both sides, \( t \approx \frac{\ln(0.5)}{\ln(0.975)} \approx 27.72 \) years.

Step 6: Estimating the Remaining Percentage After 100 Years

Substitute \( t = 100 \) into the decay formula: \( P(100) = 100(0.975)^{100} \). Calculate this to find \( P(100) \approx 7.72 \% \).

Key Concepts

Half-life calculation

Half-life is a fundamental concept in understanding exponential decay, especially in the context of radioactive materials. Simply put, half-life is the time required for a quantity to reduce to half its initial amount. In real-world terms, it's how long it takes for half of a radioactive substance to undergo decay and turn into something else.

For strontium-90, which is a byproduct of nuclear reactions, the half-life can be calculated using the formula for exponential decay. The formula is:

• \[ P(t) = P_0 (0.975)^t \]

This represents the percentage of strontium-90 remaining after time \( t \), given a decay rate of 2.5\% per year.

To find the half-life, you set \( P(t) = 50 \) in this formula, because you are looking for the time when only 50\% of the original amount is left. By solving the equation \[ 0.5 = (0.975)^t \] using logarithms, you find that the half-life of strontium-90 is approximately 27.72 years. This means after this time frame, half of the initial strontium-90 has decayed.

Radioactive decay

Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. In the case of strontium-90, as it decays, it transforms into another element. This process follows an exponential decay model.This model is represented by the equation \[ P(t) = P_0 e^{-kt} \], where \( P(t) \) is the remaining substance at time \( t \), \( P_0 \) is the initial amount, and \( k \) is the decay constant.

The decay constant is found using \[ k = ext{ln}(1 - ext{decay rate}) \]. For strontium-90, with a decay rate of 2.5\%, you calculate \( k = ext{ln}(0.975) \).

This constant \( k \) shows how quickly or slowly the substance is decaying. A higher value of \( k \) would indicate a faster decay process. Understanding this constant is crucial for making accurate predictions about the material's future activity.

Graphing exponential functions

Graphing exponential functions provides a visual representation of how quantities change over time. In the case of radioactive decay, such as the strontium-90 example, graphing provides insights into how the remaining percentage of a substance decreases.The function \[ P(t) = 100(0.975)^t \] describes the decay over time, starting completely intact at 100\% (\( P_0 = 100 \)). When graphed, this function forms a downward-sloping curve which gets closer to zero but never actually reaches it as time progresses.

Plotting \( P \) against \( t \) from time \( t=0 \) to 100 years, you would observe:

• A rapid decline at the start of the curve
• A slowing rate of decline as \( t \) increases
• After about 100 years, as per calculations, a small portion (around 7.72\%) of the original substance remains

This graphing approach helps illustrate the concept of half-life and the nature of exponential decay, making it easier to comprehend how radioactive substances diminish over time.