Question

Strontium-90 is a radioactive material that decays according to the function \(A(t)=A_{0} e^{-0.0244 t}\) where \(A_{0}\) is the initial amount present and \(A\) is the amount present at time \(t\) (in years). Assume that a scientist has a sample of 500 grams of strontium-90. (a) What is the decay rate of strontium-90? (b) How much strontium-90 is left after 10 years? (c) When will 400 grams of strontium-90 be left? (d) What is the half-life of strontium-90?

Short answer

Decay rate: 0.0244. After 10 years: 392.41 grams. 400 grams left after 9.14 years. Half-life: 28.41 years.

Step-by-step solution

Identify the decay rate

The decay rate of a radioactive material is given directly in the exponent of the decay function. The function provided is \[ A(t) = A_{0} e^{-0.0244 t} \].Thus, the decay rate is 0.0244.

Calculate the amount left after 10 years

To find how much strontium-90 is left after 10 years, use the provided decay formula. Set \( t = 10 \) and \( A_{0} = 500 \): \[ A(10) = 500 e^{-0.0244 \times 10} \].Simplify the exponent: \[ e^{-0.244} \].Evaluate:\[ A(10) = 500 \times e^{-0.244} \approx 500 \times 0.784819 \approx 392.41 \text{ grams} \].

Determine when 400 grams will be left

Set \( A(t) = 400 \) and solve for \( t \): \[ 400 = 500 e^{-0.0244 t} \].Divide both sides by 500: \[ \frac{400}{500} = e^{-0.0244 t} \].Simplify the fraction: \[ 0.8 = e^{-0.0244 t} \].Take the natural logarithm of both sides: \[ \text{ln}(0.8) = -0.0244 t \].Solve for \( t \): \[ t = \frac{\text{ln}(0.8)}{-0.0244} \approx \frac{-0.2231}{-0.0244} \approx 9.14 \text{ years} \].

Find the half-life of strontium-90

The half-life is the time it takes for half of the initial amount to remain. Set the equation as follows for half-life (\( t_{1/2} \)): \[ \frac{A_{0}}{2} = A_{0} e^{-0.0244 t_{1/2}} \].Divide both sides by \( A_{0} \): \[ \frac{1}{2} = e^{-0.0244 t_{1/2}} \].Take the natural logarithm of both sides: \[ \text{ln}\frac{1}{2} = -0.0244 t_{1/2} \].Solve for \( t_{1/2} \): \[ t_{1/2} = \frac{\text{ln}\frac{1}{2}}{-0.0244} = \frac{-0.6931}{-0.0244} \approx 28.41 \text{ years} \].

Key concepts

Exponential Decay

Exponential decay is a type of decay process in which the quantity of a material decreases by a constant percentage rate over equally spaced time intervals. It is typically represented by the equation \[ A(t) = A_{0} e^{-kt} \]. Here, \( A(t) \) is the amount remaining at time \( t \), \( A_{0} \) is the initial amount, \( e \) is the base of the natural logarithm, and \( k \) is the decay constant. Exponential decay is often seen in radioactive materials, chemicals in reactions, and some population dynamics.
In our exercise, Strontium-90 decays exponentially, as given by the function \[ A(t) = A_{0} e^{-0.0244 t} \]. The constant \( 0.0244 \) is the decay rate, determining the rate at which the substance decreases. This constant directly influences how quickly Strontium-90 decays over time.
The advantages of understanding exponential decay include predicting the remaining quantity of a substance after a certain period, knowing how long it takes for a substance to decrease to a specific amount, or determining the half-life of a material.

Half-Life Calculation

The half-life of a radioactive substance is the time it takes for half of the initial amount to decay. It is a critical estimation in understanding how quickly a radioactive material loses its potency. To find the half-life (\( t_{1/2} \)), we use the equation \[ \frac{A_{0}}{2} = A_{0} e^{-kt_{1/2}} \].
Simplifying, it becomes: \[ \frac{1}{2} = e^{-kt_{1/2}} \]. Taking the natural logarithm (ln) of both sides, we get: \[ \text{ln}\frac{1}{2} = -kt_{1/2} \].
For Strontium-90, substituting \( k = 0.0244 \), we calculate: \[ t_{1/2} = \frac{\text{ln}\frac{1}{2}}{-0.0244} = \frac{-0.6931}{-0.0244} \ \rightarrow t_{1/2} \text{ is approximately 28.41 years} \]. This means it takes about 28.41 years for half of the initial amount of Strontium-90 to decay.

Decay Rate Determination

The decay rate of a substance is a measure of how quickly it loses mass or activity over time. In exponential decay, it is represented by the constant \( k \) in the decay equation \[ A(t) = A_{0} e^{-kt} \].
A higher decay rate constant \( k \) indicates faster decay, while a lower \( k \) indicates slower decay. For Strontium-90, the decay function given is \[ A(t) = A_{0} e^{-0.0244 t} \], which directly tells us that the decay rate is \( 0.0244 \text{ per year} \).
The decay rate can be used to calculate how much of the substance will remain after a given period and to determine the half-life of the material. It provides a valuable tool for scientists and engineers to predict the behavior of the substance over time.

Natural Logarithm

The natural logarithm, denoted as \( \text{ln} \), is a logarithm to the base \( e \), where \( e \approx 2.71828 \). It is widely used in various branches of mathematics and science, including in exponential decay problems.
When solving exponential decay equations, taking the natural logarithm of both sides simplifies solving for time \( t \). For instance, if we need to find the time when a material decays to a particular amount, we use \( \text{ln} \) to isolate \( t \).
In our exercise, to find when 400 grams of Strontium-90 are left, we start with: \[ 400 = 500 e^{-0.0244 t} \], which simplifies to: \[ \frac{400}{500} = e^{-0.0244 t} \]. Taking the natural logarithm of both sides gives: \[ \text{ln}(0.8) = -0.0244 t \]. Solving for \( t \): \[ t = \frac{\text{ln}(0.8)}{-0.0244} \ \rightarrow t \text{ is approximately 9.14 years} \].
The natural logarithm helps simplify and solve equations that involve exponential functions, making it a crucial tool in understanding exponential decay.