Problem 73 - Algebra and Trigonometry with Analytic Geometry

Question

Radioactive iodine \({ }^{131} \mathrm{I}\) is frequently used in tracer studies involving the thyroid gland. The substance decays according to the formula \(A(t)=A_{0} a^{-t}\), where \(A_{0}\) is the initial dose and \(t\) is the time in days. Find \(a\), assuming the half-life of \({ }^{131} \mathrm{I}\) is 8 days.

Step-by-Step Solution

Answer

The value of \(a\) is \(2^{1/8}\).

Step 1: Understanding the Problem

The problem involves a radioactive substance that decays with time. We are given the formula \(A(t) = A_0 a^{-t}\). Here, \(A_0\) is the initial amount, and \(t\) is time. Our task is to find the value of \(a\) given that the half-life of \(^{131} \mathrm{I}\) is 8 days.

Step 2: Definition of Half-Life

The half-life of a substance is the time required for half of the initial amount to remain. Mathematically, when \(t\) equals the half-life, \(A(t) = \frac{A_0}{2}\). Therefore, the equation becomes \(\frac{A_0}{2} = A_0 a^{-8}\) when the half-life is 8 days.

Step 3: Simplifying the Equation

First, divide both sides of the equation \(\frac{A_0}{2} = A_0 a^{-8}\) by \(A_0\) (assuming \(A_0 eq 0\)) to get \(\frac{1}{2} = a^{-8}\).

Step 4: Solving for \(a\)

To isolate \(a\), take the reciprocal of both sides, which gives \(2 = a^8\). Next, take the eighth root on both sides to solve for \(a\), yielding \(a = 2^{1/8}\).

Step 5: Concluding with the Result

After simplifying, we find that the base of the exponential decay function, \(a\), is \(2^{1/8}\). This matches the given half-life condition.

Key Concepts

Half-Life

Half-life is a crucial concept when studying radioactive decay. It refers to the time required for a substance to reduce to half of its initial amount. For instance, if you start with 100 grams of a radioactive material, after one half-life, you will have 50 grams left. This process continues, halving the amount with each passing half-life.

In the context of the exercise, the half-life of iodine-131 is given as 8 days. This means, every 8 days, the amount of iodine-131 will be half of what it was at the beginning of that period.

The mathematical expression used to represent half-life in decay problems is crucial. When time \( t \) is equal to the half-life, the amount left \( A(t) \) should be half the initial amount \( \left(\frac{A_0}{2}\right) \). This gives us the foundation to resolve equations involving exponential decay.

Radioactive Decay

Radioactive decay is a natural process where unstable atomic nuclei lose energy by emitting radiation. This process continues until the unstable atoms transform into a more stable form. Each radioactive material has its own decay rate, measured in terms of half-life.

During decay, a characteristic pattern emerges - the quantity of radioactive material decreases exponentially over time. For iodine-131, the half-life of 8 days means that the activity or the measurable decay quantity, described by \( A(t) \), follows a predictable exponential pattern.

• The decay is often modelled using formulas like \( A(t) = A_0 a^{-t} \).
• In this formula, \( A_0 \) is the initial amount of the substance, and \( t \) represents time in days.
• The base \( a \) of the exponential is crucial in determining how quickly the substance decays.

Understanding radioactive decay helps in various applications such as medicine and archeology, providing a timeline for how long materials need to be managed.

Algebraic Equations

Algebraic equations are used to solve problems by finding unknown variables. In our exercise, the equation \( \frac{A_0}{2} = A_0 a^{-8} \) helps find the base of the exponential decay. Here's how the equation is simplified and solved:

Initially, divide the equation by \( A_0 \) assuming it's not zero, leading to \( \frac{1}{2} = a^{-8} \). This simplifies the problem into solving for \( a \) without variables that might seem confusing.

• To solve \( \frac{1}{2} = a^{-8} \), rewrite it by taking reciprocal, resulting in \( 2 = a^8 \).
• Next, to isolate \( a \), take the eighth root of both sides, yielding \( a = 2^{1/8} \).

This example highlights how algebra can simplify even complicated problems, making them manageable and easier to understand. Solving such equations requires logical thinking and familiarity with algebraic manipulation techniques, crucial skills in both academic and real-world contexts.