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Chapter 4: Problem 395

A substance has a half-life of 2.045 minutes. If the initial amount of the substance was 132.8 grams, how many half-lives will have passed before the substance decays to 8.3 grams? What is the total time of decay?

Short Answer

Expert verified
Approximately 4 half-lives and total decay time is 8.18 minutes.

Step by step solution

01

Understanding the Half-Life Concept

The half-life of a substance is the time it takes for half of the substance to decay. We are given a half-life of 2.045 minutes.
02

Finding Number of Half-Lives

We need to determine how many half-lives pass for the initial mass of 132.8 grams to decay to 8.3 grams. During each half-life, the substance reduces to half of its previous amount.
03

Using the Half-Life Formula

The relationship of the amount of decay can be modeled as: \\( A = A_0 \times \left(\frac{1}{2}\right)^n \) \where \( A_0 = 132.8 \) grams, final amount \( A = 8.3 \) grams, and \( n \) is the number of half-lives.
04

Solving for n

Substitute the values into the equation: \\( 8.3 = 132.8 \times \left(\frac{1}{2}\right)^n \) \Divide both sides by 132.8: \\( \frac{8.3}{132.8} = \left(\frac{1}{2}\right)^n \). \Take the logarithm of both sides to isolate \( n \): \\( n = \frac{\log\left(\frac{8.3}{132.8}\right)}{\log\left(\frac{1}{2}\right)} \).
05

Calculate Number of Half-Lives

Calculate \( n \): \\( n \approx \frac{\log(0.0625)}{\log(0.5)} \approx 4 \). \Thus, approximately 4 half-lives have passed.
06

Calculate Total Time of Decay

Multiply the number of half-lives by the duration of each half-life to find the total decay time: \\[ \text{Total time} = n \times \text{half-life} = 4 \times 2.045 \].
07

Final Calculation

\( \text{Total time} = 8.18 \text{ minutes} \).

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Key Concepts

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

Radioactive Decay
Radioactive decay is a natural process through which an unstable atomic nucleus loses energy by radiation. It's a fascinating and fundamental concept in physics that describes how unstable isotopes transform over time. These transformations occur at a characteristic rate for each type of isotope, which we call "decay rate." This process is random for any single atom, but predictable for a large number of nuclei.
  • Over a period of time, a certain number of atoms will always decay, following the principles of probability.
  • This process continues until the material transforms into a stable state and can produce various particles, such as alpha particles or beta particles.
Radioactive decay is key to understanding the behavior of radioactive substances in various settings. It helps explain how isotopes are used in medicine, archaeology, and other fields.
Logarithms
Logarithms are a mathematical tool used to solve equations involving exponential growth or decay, like those found in radioactive decay problems. Essentially, a logarithm tells us how many times we need to multiply a specific number (known as the "base") to reach another number.
  • For example, what power do we raise 10 to obtain 100? The answer is 2, because \(10^2 = 100\). Thus, \( ext{log}_{10}(100)=2\).
  • In the context of half-life calculations, the logarithm helps determine how many half-lives have passed. The base here is often 2, reflecting the nature of the halving process.
Using logarithms simplifies the process of unwinding exponential equations. In our exercise, applying logarithms to solve for "n," the number of half-lives, lets us find the solution efficiently.
Exponential Decay
Exponential decay is a type of reduction that occurs at a consistent rate over time. This principle is crucial in describing processes like radioactive decay, where the quantity of a substance decreases rapidly at first and then levels off more slowly.
  • The formula to represent exponential decay is \( A = A_0 imes ext{e}^{-kt} \), where \( A_0 \) is the initial amount, \( A \) is the amount at time \( t \), \( k \) is the decay constant, and \( e \) is the base of the natural logarithm.
  • In radioactive decay, this is simplified since the decay occurs in set half-lives, using powers of \( \frac{1}{2} \) instead of \( e \).
Exponential decay results in a curve that gets closer to zero asymptotically, which means it never truly reaches zero but declines steadily. This concept forms the backbone of understanding how radioactive substances decrease over time.
Half-Life Formula
The half-life formula is a simplified way to compute the decay of substances over time, tailored specifically for processes where a substance halves at regular intervals. This notion is thoroughly practical in many fields, including geology, archaeology, and physics.
  • Mathematically, the formula is expressed as: \( A = A_0 \times \left(\frac{1}{2}\right)^n \), where \( A \) is the final amount, \( A_0 \) is the initial amount, and \( n \) is the number of half-lives that have passed.
  • For our problem, finding \( n \) involved rearranging the formula and using logarithms to solve for the number of half-lives needed for the substance to decrease from 132.8 grams to 8.3 grams.
Knowing how to use and interpret the half-life formula allows us to predict future states of a substance undergoing radioactive decay, making it an invaluable tool for scientists and researchers.

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Most popular questions from this chapter

For the following exercises, refer to Table 4.26. $$\begin{array}{|c|c|c|c|c|c|c|}\hline x & {1} & {2} & {3} & {4} & {5} & {6} \\\ \hline f(x) & {1125} & {1495} & {2310} & {3294} & {4650} & {6361} \\\ \hline\end{array}$$ Write the exponential function as an exponential equation with base \(e\).

Determine whether the function\(y=156(0.825)^{t}\) represents exponential growth, exponential decay, or neither. Explain

For the following exercises, use properties of logarithms to evaluate without using a calculator. $$ 6 \log _{8}(2)+\frac{\log _{8}(64)}{3 \log _{8}(4)} $$

For the following exercises, refer to Table 4.31. $$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline x & {0} & {2} & {4} & {5} & {7} & {8} & {10} & {11} & {15} & {17} \\ \hline f(x) & {12} & {28.6} & {52.8} & {70.3} & {99.9} & {112.5} & {125.8} & {127.9} & {135.1} & {135.9} \\\ \hline\end{array}$$ Use the intersect feature to find the value of \(x\) for which the model reaches half its carrying capacity.

Use logarithms to solve. \(7 e^{3 n-5}+5=-89\)
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