Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

If a radioactive substance loses \(15 \%\) of its radioactivity in 2 days, what is its half-life?

Step-by-Step Solution

Answer

The half-life is approximately 8.87 days.

Step 1: Understand the Problem

We are given that a radioactive substance loses 15% of its radioactivity in 2 days, and we are asked to find its half-life. To find the half-life, we will need to use the concept of exponential decay and logarithms.

Step 2: Set Up the Exponential Decay Formula

The formula for exponential decay is given by \( N(t) = N_0 \, e^{-kt} \), where \( N(t) \) is the amount of substance at time \( t \), \( N_0 \) is the initial amount, \( k \) is the decay constant, and \( t \) is the time. Since the substance loses 15% of its original radioactivity in 2 days, after 2 days, we have \( N(t) = 0.85 N_0 \).

Step 3: Solve for the Decay Constant (k)

Substitute the given into the decay formula: \( 0.85 N_0 = N_0 e^{-2k} \).Cancel \( N_0 \) from both sides: \( 0.85 = e^{-2k} \).Take the natural logarithm of both sides: \( \ln(0.85) = -2k \).Solve for \( k \): \( k = -\frac{\ln(0.85)}{2} \).

Step 4: Apply the Half-life Formula

The formula for half-life \( T_{1/2} \) is \( T_{1/2} = \frac{\ln(2)}{k} \). We will substitute the value of \( k \) from Step 3 here.

Step 5: Calculate the Half-life

Using the value of \( k = -\frac{\ln(0.85)}{2} \), set \( T_{1/2} = \frac{\ln(2)}{-\frac{\ln(0.85)}{2}} = \frac{2 \ln(2)}{-\ln(0.85)} \). Calculate the value using a calculator to find the numerical solution.

Key Concepts

Exponential Decay

Exponential decay is a process where a quantity decreases at a rate proportional to its current value. This means the greater the amount, the larger the decrease in a fixed period. This smooth and rapid decrease described by exponential functions can be seen in numerous natural and human-made processes.

For a clearer understanding, consider the exponential decay formula:

• \( N(t) = N_0 \, e^{-kt} \), where:
  • \( N(t) \) is the quantity at time \( t \)
  • \( N_0 \) is the initial quantity
  • \( k \) is the decay constant
  • \( t \) is time

This formula illustrates how a constant fraction of the substance disappears in each time period. Notably, the decay constant \( k \) plays a crucial role in determining the rate of decay: the larger the 'k', the quicker the decay.

When we apply this to problems such as calculating the half-life of a radioactive material, understanding the role of \( k \) and \( e \), the base of natural logarithms, is vital for solving the problem efficiently.

Radioactive Decay

Radioactive decay refers to the natural decrease of an unstable atomic nucleus. When an unstable atom transforms into a more stable form, it releases energy in the form of radiation. This change is both spontaneous and irreversible, and it follows the pattern of exponential decay.

Key aspects include:

• Each type of radioactive material has a unique decay rate.
• This process results in a loss of mass, energy, or both.
• The "half-life" is the time taken for half the substance to decay.

To handle these calculations, scientists use formulas derived from the general exponential decay formula, substituting in specific values that describe the material's decaying properties. Understanding radioactive decay is crucial not only in physics but also in various fields like medicine and archaeology, where radioactive substances are used for treatments and dating ancient artifacts.

Logarithms in Calculus

Logarithms in calculus help solve exponential equations by transforming complex multiplicative processes into simpler additive ones. They are particularly handy in solving exponential decay problems like half-life calculations.

The natural logarithm, denoted as \( \ln \), is a critical tool as it uses base \( e \). Here's why it's useful:

• It allows conversion of exponential forms into linear ones, making equations easier to handle, for example: \[ e^x = y \Longrightarrow x = \ln(y) \]
• Provides an efficient method to determine decay constants by isolating \( k \) in decay equations.

By applying logarithms, you can methodically solve for variables that may not be easily accessible otherwise. It simplifies the complex processes inherent in exponential decay calculations and supports finding exact solutions in half-life problems. Understanding and applying logarithms in calculus is, therefore, indispensable for mastering topics in math and science that involve decay processes.