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Chapter 5: Problem 16

Estimate the time it will take an initial quantity to drop to half its value when: a. \(P=3.02(0.998)^{t}\), with \(t\) in years b. \(Q=12(0.75)^{T}\), with \(T\) in decades

Short Answer

Expert verified
Part (a): Approximately 346.57 years. Part (b): Approximately 1.71 decades.

Step by step solution

01

Identify the half-life formula

The half-life formula for exponential decay processes is derived from the general decay formula. For a quantity that decays to half its initial value, set the new quantity to be half of the original and solve for time.
02

Part (a): Set up the decay equation

Given the equation for the quantity: \[ P = 3.02(0.998)^t \] To find the half-life, set the equation to half of the initial value: \[ \frac{3.02}{2} = 3.02(0.998)^t \]
03

Part (a): Solve for half-life

Divide both sides by 3.02: \[ \frac{1}{2} = (0.998)^t \] Take the natural logarithm of both sides to solve for \( t \): \[ \text{ln}\frac{1}{2} = t \times \text{ln}(0.998) \] \[ t = \frac{\text{ln}(0.5)}{\text{ln}(0.998)} \] Calculate \( t \): \[ t \approx 346.57 \text{ years} \]
04

Part (b): Set up the decay equation

Given the equation for the quantity: \[ Q = 12(0.75)^T \] To find the half-life, set the equation to half of the initial value: \[ \frac{12}{2} = 12(0.75)^T \]
05

Part (b): Solve for half-life

Divide both sides by 12: \[ \frac{1}{2} = (0.75)^T \] Take the natural logarithm of both sides to solve for \( T \): \[ \text{ln}\frac{1}{2} = T \times \text{ln}(0.75) \] \[ T = \frac{\text{ln}(0.5)}{\text{ln}(0.75)} \] Calculate \( T \): \[ T \approx 1.71 \text{ decades} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

half-life calculation
To determine how long it takes for a quantity to reduce to half its initial value, we use the concept of half-life. The half-life is the time period over which the quantity decays to 50% of its original amount. For any exponential decay process, we typically use the given decay equation. For example, in part (a) of the exercise, we use the decay formula: 3.02(0.998)^t. To find the half-life, we set the final quantity to be half of the initial quantity: \[\frac{3.02}{2} = 3.02(0.998)^t\]. This equation can then be solved to find the value of t that represents the half-life.
natural logarithm
The natural logarithm, denoted as \(\text{ln}\), is a special type of logarithm that uses the constant 'e' (approximately 2.71828) as the base. It is extremely useful for solving exponential decay equations because it allows us to linearize the exponential relationship. When solving for the half-life, like in our exercise, we used the natural logarithm to make the equations simpler. For instance, in part (a) of our problem, we transformed the equation \(\frac{1}{2} = (0.998)^t\) by taking the natural logarithm of both sides. This made it possible to isolate t:\[\text{ln}\frac{1}{2} = t \times \text{ln}(0.998)\]. By solving this, we can find the half-life.
exponential functions
An exponential function is a mathematical expression in the form \(f(t) = a \cdot b^t\), where:
  • a is the initial amount,
  • b is the base of the exponential,
  • t is the variable, often representing time.
In the context of decay, the base b is a fraction less than 1. This indicates that the quantity is decreasing over time. For example, part (a) of the exercise has the exponential function 3.02(0.998)^t, which shows that the initial amount is 3.02 and it decays at a rate represented by 0.998 per unit time. Understanding how these components work together helps us solve for various parameters such as half-life.
decay equations
Decay equations describe how a quantity decreases over time. They are typically written in the form: \[Q = Q_0 \cdot b^t\], where Q is the final quantity, Q_0 is the initial quantity, b is the base (decay factor), and t is time. In the exercise, we worked with two decay equations:
  • Part (a):\[P = 3.02(0.998)^t\],
  • Part (b): \[Q = 12(0.75)^T\].
In each case, to find the half-life, we set the final quantity to half of the initial quantity: \[\frac{Q_0}{2} = Q_0 \cdot b^t\], and then solve for t, the time at which this occurs.
solving for time
Solving for time in exponential decay equations requires isolating the time variable. This is usually done by taking the natural logarithm of both sides of the equation. For example, let's look at part (a) again: Starting with: \[\frac{3.02}{2} = 3.02(0.998)^t\], we simplify to: \[\frac{1}{2} = (0.998)^t\]. Taking the natural logarithm of both sides results in: \[\text{ln}\frac{1}{2} = t \cdot \text{ln}(0.998)\]. Finally, solving for t gives us: \[t = \frac{\text{ln}(0.5)}{\text{ln}(0.998)}\]. This process reveals the half-life by finding the time at which the quantity is halved.

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