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Chapter 1: Problem 42

A certain moon rock was found to contain equal numbers of potassium and argon atoms. Assume that all the argon is the result of radioactive decay of potassium (its half-life is about \(1.28 \times 10^{9}\) years) and that one of every nine potassium atom disintegrations yields an argon atom. What is the age of the rock, measured from the time it contained only potassium?

Short Answer

Expert verified
The age of the moon rock is approximately 1.28 billion years.

Step by step solution

01

Understand the Problem

You need to find the age of a moon rock that has equal numbers of potassium and argon atoms by considering radioactive decay. Initially, the rock contained only potassium and over time, some of it decayed to argon.
02

Set Up the Decay Equation

Use the decay formula \( N(t) = N_0 e^{-\lambda t} \) where \( N(t) \) is the number of potassium atoms remaining after time \( t \). Let \( N_0 \) be the initial number of potassium atoms and \( \lambda \) be the decay constant.
03

Relate Potassium and ArgonAtoms

Since initially there was only potassium, the initial number of potassium atoms \( N_0 \) equals the current number of potassium atoms plus argon atoms. Since there are equal numbers of each, \( N_0 = N(t) + N(t) = 2N(t) \) because every 9 disintegrations of potassium yield 1 argon atom.
04

Calculate the Decay Constant

The decay constant \( \lambda \) is related to the half-life by the formula \( \lambda = \frac{\ln(2)}{t_{1/2}} \). So, \( \lambda = \frac{\ln(2)}{1.28 \times 10^9} \).
05

Solve for Time

By substituting \( N_0 = 2N(t) \) into the decay equation, we get \( 2N(t) = N_0 e^{-\lambda t} \). Simplifying, \( ln(2) = \lambda t \). Solve for \( t \) using \( t = \frac{\ln(2)}{\lambda} \) and the value of \( \lambda \) calculated previously.
06

Compute the Age of the Rock

Compute \( t = \frac{\ln(2)}{\frac{\ln(2)}{1.28 \times 10^9}} \). This gives \( t = 1.28 \times 10^9 \) years.

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Key Concepts

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

Half-Life Calculation
Radioactive decay is a fascinating process, and one of its key aspects is the concept of half-life. Half-life is the duration needed for half of a radioactive material to transform into another element through decay.
It's vital for determining the age of rocks and fossils by examining isotopes. The half-life does not change over time or with changes in physical conditions.
  • To calculate the half-life, a relationship between time, decay constant, and half-life is used.
  • Given as: \(t_{1/2} = \frac{\ln(2)}{\lambda}\), where \ \lambda \ is the decay constant.
The half-life of potassium, for instance, is around \(1.28 \times 10^9\) years. This means it takes this long for half of a sample of potassium to decay into a different element, like argon in the case of potassium-argon dating.
Potassium-Argon Dating
One intriguing application of radioactive decay is potassium-argon dating. This technique dates geological and archaeological samples based on the decay of potassium-40 to argon-40.
  • The key idea is that as potassium in a rock decays over time, argon builds up since some potassium disintegrations convert to argon.
  • By measuring the ratio of remaining potassium to produced argon, scientists can determine how much time has passed since the sample was trapped in the rock.
This method is particularly useful for dating ancient rocks due to the long half-life of potassium-40. Therefore, when a moon rock shows equal amounts of both, it signifies that a significant period has passed since its formation.
Decay Constant
The decay constant is crucial for understanding the rate of decay of a radioactive substance. It's a proportional factor in the decay equation that helps compute how quickly a substance undergoes radioactive decay.
  • Represented by lambda \(\lambda\), the decay constant is linked with the half-life, making it important for radioactive dating calculations.
  • It is calculated through: \(\lambda = \frac{\ln(2)}{t_{1/2}}\).
When dealing with long-lived isotopes such as potassium, which has a very long half-life, the decay constant becomes a small number.
Knowing the decay constant allows scientists to solve the decay equation and determine the elapsed time since the formation of the sample through its isotopic changes.

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Most popular questions from this chapter

Write-in the manner of Eqs. (3) through (6) of this section-a differential equation that is a mathematical model of the situation described. The time rate of change of the velocity \(v\) of a coasting motorboat is proportional to the square of \(v\).

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An accident at a nuclear power plant has left the surrounding area polluted with radioactive material that decays naturally. The initial amount of radioactive material present is 15 su (safe units), and 5 months later it is still 10 su. (a) Write a formula giving the amount \(A(t)\) of radioactive material (in su) remaining after \(t\) months. (b) What amount of radioactive material will remain after 8 months? (c) How long-total number of months or fraction thereof-will it be until \(A=1 \mathrm{su}\), so it is safe for people to return to the area?

A function \(y=g(x)\) is described by some geometric property of its graph. Write a differential equation of the form \(d y / d x=f(x, y)\) having the function \(g\) as its solution (or as one of its solutions). The slope of the graph of \(g\) at the point \((x, y)\) is the sum of \(x\) and \(y\).

Problems deal with a shallow reservoir that has a one-square-kilometer water surface and an average water depth of 2 meters. Initially it is filled with fresh water, but at time \(t=0\) water contaminated with a liquid pollutant begins flowing into the reservoir at the rate of 200 thousand cubic meters per month. The well-mixed water in the reservoir flows out at the same rate. Your first task is to find the amount \(x(t)\) of pollutant (in millions of liters) in the reservoir after 1 months. The incoming water has a pollutant concentration of \(c(t)=10\) liters per cubic meter \(\left(\mathrm{L} / \mathrm{m}^{3}\right)\). Verify that the graph of \(x(t)\) resembles the steadily rising curve in Fig. 1.5.9, which approaches asymptotically the graph of the equilibrium solution \(x(t)=20\) that corresponds to the reservoir's long-term pollutant content. How long does it take the pollutant concentration in the reservoir to reach \(10 \mathrm{~L} / \mathrm{m}^{3} ?\)
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