Engaging with the intricacies of radioactive decay in chemistry often leads students through a myriad of calculations and conceptual understanding. Among these challenges, solving half-life worksheet problems is one that requires both a grasp of the underlying principles and a methodical approach. This guide delves into the heart of these challenges, offering five proven answers to common half-life worksheet problems, thereby enhancing your understanding and proficiency in radioactive decay calculations.
Understanding the Concept of Half-life
Before diving into problem-solving, a clear understanding of what half-life means is crucial:
- Half-life is the time it takes for half of the radioactive nuclei in a sample to decay.
- The decay is exponential, meaning it doesn't depend on the initial quantity of the material.
- The equation for half-life is t_{1/2} = \frac{\ln(2)}{\lambda} , where \lambda is the decay constant.
Example 1: Finding the Half-life
Given a sample of phosphorus-32 where 12.5% of the original substance remains after 28.5 days, find the half-life:
| Time (t) | Remaining (%) | No. of Half-lives (n) |
|---|---|---|
| 28.5 days | 12.5% | 3 |
Since 12.5% is equivalent to \left(\frac{1}{2}\right)^3 of the original sample:
- The number of half-lives (n) is 3, as 12.5\% = \left(\frac{1}{2}\right)^3 .
- Half-life (t1/2) can be calculated as t_{1/2} = \frac{t}{n} , where t is the given time. Hence, t_{1/2} = \frac{28.5}{3} \approx 9.5 days.
Example 2: Decayed Activity
A sample of Strontium-90 has an initial activity of 400 counts per minute. After 4 years, its activity is reduced to 200 counts per minute. Calculate the half-life:
- Calculate the number of half-lives passed using the formula for remaining activity: \text{Remaining Activity} = \text{Initial Activity} \times (\frac{1}{2})^n .
- Here, 200 counts = 400 counts \times \left(\frac{1}{2}\right)^n. So, \left(\frac{1}{2}\right)^n = \frac{200}{400} = 0.5 , which means n = 1.
- Given the activity decayed in 4 years, the half-life of Strontium-90 is t_{1/2} = \frac{4}{1} = 4 years.
🔍 Note: Strontium-90 is commonly used as an example in half-life problems due to its relatively known decay properties.
Example 3: Initial Quantity Calculation
You are given that after 3 half-lives, a sample has decayed to 25% of its original amount. What is the initial amount if the remaining amount is 16 grams?
- Initial amount (N0) can be calculated using the half-life formula ( N_t = N_0 \times (\frac{1}{2})^n ), where n is the number of half-lives.
- Here, 25% = ( (\frac{1}{2})^3 ), so ( N_0 = \frac{16 \text{ grams}}{(\frac{1}{2})^3} = 64 \text{ grams} ).
Example 4: Radioactive Isotopes and Age Dating
If a rock sample containing Rubidium-87 shows that 20% of the isotope has decayed into Strontium-87, how old is the rock? (Rubidium-87 has a half-life of 48.8 billion years).
- Calculate the fraction of Rubidium-87 remaining: ( 1 - 0.20 = 0.80 ).
- Find the number of half-lives by solving ( 0.80 = \left(\frac{1}{2}\right)^n ), which leads to ( \log_{0.5}0.80 = n ). Using log properties, ( n \approx 0.321 ).
- Age of the rock = Number of half-lives (n) × Half-life. So, ( n \times 48.8 ) billion years (\approx 15.7 ) billion years.
Example 5: Using a Differential Equation
Solve for the time taken for the activity of a radioactive isotope with a decay constant of 0.115 to drop to 25% of its initial activity:
- The rate of decay equation is ( \frac{dA}{dt} = -\lambda A ), where A is activity, λ is the decay constant, and t is time.
- Integrating gives ( A_t = A_0 e^{-\lambda t} ). Since ( A_t ) is 25% of ( A_0 ), we have ( 0.25A_0 = A_0 e^{-0.115t} ).
- Solving for t, we get ( t \approx \frac{\ln(0.25)}{-0.115} \approx 15.8 \text{ years} ).
Thus, by navigating through these examples, you can observe how versatile and essential half-life calculations are in understanding radioactive decay. Each problem not only reinforces the concept of exponential decay but also provides practical applications in science and daily life. Mastery of these calculations can significantly enhance one's ability to interpret complex radioactive processes, contributing to both educational and practical advancements.
Wrapping up, these examples have showcased the broad spectrum of half-life problem-solving techniques. From calculating decay times to determining the age of artifacts, the half-life concept is a cornerstone in understanding nuclear stability and time-dependent phenomena. By honing these skills, one can better appreciate the continuity and change that governs much of the natural world.
What is the significance of half-life?
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Half-life is crucial in dating geological or biological samples, medical treatments with isotopes, and understanding nuclear decay processes. It tells us the rate at which a radioactive substance decays, helping predict the behavior of radioactive materials over time.
How can I remember the half-life formula?
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Consider the mnemonic: “Half-life to go, Lambda to know.” ( t_{1⁄2} = \frac{\ln(2)}{\lambda} ) where “to go” refers to time (t1⁄2) and “to know” refers to knowing the decay constant (λ).
What are some real-world applications of half-life?
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Half-life concepts are applied in carbon dating (age determination), radiology for treating cancer, environmental science for tracking pollution, and in nuclear power plant operations for safety and maintenance.
What are the common mistakes when solving half-life problems?
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Some frequent errors include: misinterpreting the number of half-lives, confusion between activity decay and mass decay, and misunderstanding the exponential nature of the decay process.
Can the half-life of an isotope change?
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No, the half-life of a particular isotope is a constant characteristic. It does not change with the quantity of the material or external conditions like temperature or pressure; it is inherent to the isotope itself.
