Problem 10 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Strontium \(90(\mathrm{Sr}-90)\), a radioactive isotope of strontium, is present in the fallout resulting from nuclear explosions. It is especially hazardous to animal life, including humans, because, upon ingestion of contaminated food, it is absorbed into the bone structure. Its half-life is \(27 \mathrm{yr}\). If the amount of \(\mathrm{Sr}-90\) in a certain area is found to be four times the "safe" level, find how much time must elapse before an "acceptable level" is reached.

Step-by-Step Solution

Answer

It will take approximately 54 years for the amount of Strontium-90 to reach the acceptable level.

Step 1: Set up the decay equation

First, we need to set up the decay equation that describes the amount of Strontium-90 remaining after a certain period of time has passed: \[A_t = A_0 \cdot (1/2)^{t/T}\]

Step 2: Express a relation between initial and final amounts

Since the current amount of Strontium-90 is 4 times the safe level. Let \(A_{safe}\) represent the safe level, so the initial amount of Strontium-90 is \(4A_{safe}\), and the final amount is \(A_{safe}\). Thus, we can write the relation as: \[A_{safe} = 4A_{safe} \cdot (1/2)^{t/T}\]

Step 3: Solve for t

Now we need to solve this equation for the time \(t\). First, we simplify the equation by dividing both sides by \(A_{safe}\): \[(1/2)^{t/T} = 1/4 \] Now, we take the natural logarithm of both sides: \[ \ln{(1/2)^{t/T}} = \ln{(1/4)}\] Using the property of logarithm \(\ln{x^y} = y \ln{x}\): \[ \frac{t}{T} \cdot \ln{(1/2)} = \ln{(1/4)}\] Now, divide by the \(\ln{(1/2)}\) and multiply by the half-life \(T\) on both sides: \[t = \frac{\ln{(1/4)}}{\ln{(1/2)}} \cdot T\] Finally, we substitute the value of \(T\) (half-life) i.e. \(27\) in the equation: \[ t = \frac{\ln{(1/4)}}{\ln{(1/2)}} \cdot 27\]

Step 4: Calculate t

Now, we can use a calculator to compute the value of \(t\): \[t \approx 54\] So it will take approximately 54 years for the amount of Strontium-90 to reach the acceptable level.

Key Concepts

Half-Life

The concept of half-life is essential to understanding radioactive decay. It refers to the time required for half of a radioactive substance to decay and transform into a different element or isotope. For Strontium-90, the half-life is 27 years. This means that every 27 years, only half of the existing Strontium-90 atoms will remain, with the rest having decayed into other elements.

Half-life helps us predict how long a substance will remain hazardous. Strontium-90, with a half-life of 27 years, becomes significantly less risky over time. After two half-life periods (or 54 years), only a quarter of the original amount will remain. This predictability is crucial for managing radioactive materials and understanding their long-term risks.

In summary:

• Half-life is the time taken for half of a radioactive material to decay.
• Strontium-90's half-life is 27 years.
• Knowing the half-life helps in planning safe handling and disposal of radioactive substances.

Decay Equation

The decay equation is a formula used to calculate the remaining quantity of a radioactive substance over time. It is expressed as: \[A_t = A_0 \cdot (\frac{1}{2})^{t/T}\] Here:

• \(A_t\) is the amount remaining after time \(t\).
• \(A_0\) is the initial amount of the substance.
• \(t\) is the elapsed time.
• \(T\) is the half-life.

This equation tells us how much of the radioactive isotope will be left after a certain period, taking into account its half-life. By applying this equation, we can assess when it's safe to be around a decaying radioactive material.

In this problem, Strontium-90's starting amount is four times the safe level. By rearranging the decay equation, we found how long it takes to decrease to a safe amount. This precise calculation aids decision-making regarding public safety and environmental hazards.

Logarithm Properties

Logarithms are crucial in solving decay problems. They help to simplify exponential equations and make them solvable. In the context of the decay equation, we need to use the logarithm to solve for time \(t\).

By leveraging the property \(\ln{x^y} = y \ln{x}\), we can solve for \(t\) in the decay equation. Here is how it applies:

• The decay equation \((1/2)^{t/T} = 1/4\) is first subjected to the natural logarithm on both sides.
• This transforms the equation using the property mentioned, simplifying it to \(\frac{t}{T} \cdot \ln{(1/2)} = \ln{(1/4)}\).
• By isolating \(t\), we can solve for how long it will take for the substance to reach a certain level.

In our scenario, using these logarithm properties allowed us to determine that it takes approximately 54 years for the Strontium-90 levels to become safe. This step is crucial as it translates complex radioactive decay processes into simpler calculations, aiding in planning and management of radioactive exposure.