Problem 395
Question
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?
Step-by-Step Solution
Verified Answer
Approximately 4 half-lives and total decay time is 8.18 minutes.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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.
4Step 4: 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)} \).
5Step 5: 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.
6Step 6: 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 \].
7Step 7: Final Calculation
\( \text{Total time} = 8.18 \text{ minutes} \).
Key Concepts
Radioactive DecayLogarithmsExponential DecayHalf-Life Formula
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.
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.
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 \( e \).
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.
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