Chapter 26: Problem 88
Moon Rocks In one of the rocks brought back from the Moon, \(75 \%\) of the initial potassium- 40 was found to have decayed to argon-40. (a) If the half- life for potassium- 40 is \(1.20 \times 10^{9}\) years, how old is the rock? (b) How much longer will it take before only \(6.25 \%\) of the original potassium- 40 is still present in the rock?
Short Answer
Step by step solution
Understand the Problem
Use Decay Formula
Solve for Time (t) for Part (a)
Calculate Time t for Part (a)
Calculate Additional Time for Part (b)
Solve for Time (t) for Part (b)
Calculate Additional Time Needed
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Potassium-40
This element is commonly found in rocks and minerals along with other potassium isotopes. The decay of Potassium-40 provides a reliable 'clock' that geologists use to determine the age of rock samples.
Here's why it's important:
- It helps measure geological time scales, which are crucial for understanding Earth's history.
- This isotope is important in geochronology, a branch of geology that focuses on determining the age of rocks, fossils, and sediments.
- Its slow decay rate makes it ideal for dating extremely old samples, such as lunar rocks brought back from the Moon missions.
Half-life
Understanding half-life is crucial because:
- It allows scientists to calculate the age of an object by determining how many half-lives have occurred.
- It's a constant that does not depend on the initial amount of the substance, ensuring accuracy in measurements.
- It helps in interpreting radioactive decay equations, like calculating the remaining quantity of a substance over time using the formula: \[ N = N_0 \left( \frac{1}{2} \right)^{t/T} \]
Logarithms
In the context of radioactive decay, logarithms help solve for the time when a specific fraction of a radioactive isotope remains. The decay formula \[ \frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T} \] can be rearranged using logarithms to find the time elapsed, \( t \).
Here's how they are applied:
- Taking the logarithm of both sides of the exponential equation converts it to a linear form, making it easier to solve for unknowns such as time \( t \).
- The base for the logarithm in decay equations is usually 0.5, aligning with the half-life concept.
- Using logarithms can let you quickly determine the number of half-lives that have passed, and thus the age of a sample.