Problem 21 - Precalculus Mathematics for Calculus

Question

These exercises use the radioactive decay model. If \(250 \mathrm{mg}\) of a radioactive element decays to \(200 \mathrm{mg}\) in 48 hours, find the half-life of the element.

Step-by-Step Solution

Answer

The half-life of the element is approximately 151.3 hours.

Step 1: Understanding Radioactive Decay

Radioactive decay follows the model \( A(t) = A_0 e^{-kt} \), where \( A(t) \) is the amount of substance at time \( t \), \( A_0 \) is the initial amount of substance, and \( k \) is the decay constant.

Step 2: Setting Up the Decay Equation

We know the initial amount \( A_0 = 250 \; mg \) and the remaining amount after 48 hours is \( A(48) = 200 \; mg \). Using the decay model equation: \[ 200 = 250 e^{-k \cdot 48} \]

Step 3: Solving for the Decay Constant \( k \)

Rearrange the equation to solve for \( e^{-48k} \):\[ e^{-48k} = \frac{200}{250} \] or \[ e^{-48k} = 0.8 \].Take the natural logarithm on both sides: \[ -48k = \ln(0.8) \]Solve for \( k \):\[ k = -\frac{\ln(0.8)}{48} \approx 0.004579. \]

Step 4: Using Half-life Formula

The half-life \( T_{1/2} \) is the time it takes for the substance to decay to half its initial amount. The formula for half-life is: \[ T_{1/2} = \frac{\ln(2)}{k} \].Substitute \( k \) to find \( T_{1/2} \):\[ T_{1/2} = \frac{\ln(2)}{0.004579} \].

Step 5: Calculating the Half-life

Compute the half-life using the values:\[ T_{1/2} \approx \frac{0.693}{0.004579} \approx 151.3 \; \text{hours}. \]

Key Concepts

Decay Constant

In the study of radioactive decay, the decay constant, denoted by the symbol \( k \), plays a crucial role. It is a measure that represents the probability per unit time that a particular atom will decay. The decay constant helps in modeling how quickly a radioactive substance decreases over time.

• The decay constant is found in the decay model equation: \[ A(t) = A_0 e^{-kt} \]
• Here, \( A(t) \) is the amount of the substance at time \( t \), \( A_0 \) is the initial amount, and \( k \) is the decay constant.

It determines the rate at which the amount of substance decreases. The greater the decay constant, the faster the substance decays. To calculate \( k \), you need to set up an equation based on known values of the substance at two different times, as in the worked-out example. This involves solving an exponential equation, often by taking the natural logarithm to solve for \( k \). Learning how to do this will help you make predictions about how long a substance will take to decay to a specific level.

Half-life Calculation

The half-life of a radioactive element is a critical concept in understanding decay processes. It is defined as the time required for a quantity to reduce to half its initial value. The formula used to calculate half-life \( T_{1/2} \) involves the decay constant \( k \):\[T_{1/2} = \frac{\ln(2)}{k}\] This formula emerges from the nature of the exponential decay described by the decay model. Here's how it works:

• Since the decay model follows an exponential curve, half-life remains constant over time.
• The half-life is independent of the initial amount of the substance.

When you compute the half-life from the decay constant, it's crucial because it informs you about the timeframe over which the substance reduces to a manageable amount. This is particularly important in fields such as archaeology, medicine, and nuclear physics, where understanding radioactive decay informs practical applications.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is an essential mathematical concept often used in solving problems involving exponential functions, such as radioactive decay. The natural logarithm has a base of \( e \), the Euler's number, approximately equal to 2.718.

• Using \( \ln \) helps convert exponential equations into linear ones, simplifying the process of solving them.
• In contexts like radioactive decay, natural logarithm enables us to solve for unknowns, such as the decay constant.

In problems where you have equations of the form \( e^x = a \), taking the natural logarithm of both sides yields \( x = \ln(a) \), effectively isolating \( x \). In the exercise example:

• We transformed the expression \( e^{-48k} = 0.8 \) by taking \( \ln(0.8) \), allowing us to solve for the decay constant \( k \).

Understanding natural logarithms is key to tackling and simplifying the mathematical intricacies involved in the study of exponential growth and decay, making them invaluable in scientific calculations.