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Chapter 9: Problem 27

A rock containing \(1 \mathrm{mg}\) of plutonium- 239 per \(\mathrm{kg}\) of rock is found in a glacier. The half-life of plutonium-239 is 25,000 years. If this rock was deposited 100,000 years ago during an ice age, how much plutonium-239 per kilogram of rock was in the rock at that time?

Short Answer

Expert verified
16 mg/kg of plutonium-239 was in the rock 100,000 years ago.

Step by step solution

01

Understanding the Problem

We need to calculate the initial amount of plutonium-239 that was present 100,000 years ago given that the current amount is 1 mg/kg. We know the half-life of plutonium-239 is 25,000 years.
02

Apply the Half-life Formula

Use the formula for exponential decay: \( N(t) = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \), where \( N(t) \) is the amount remaining after time \( t \), \( N_0 \) is the initial amount, and \( T_{1/2} \) is the half-life.
03

Calculate Number of Half-lives

Find the number of half-lives that have passed in 100,000 years: \( \frac{t}{T_{1/2}} = \frac{100,000}{25,000} = 4 \).
04

Calculate Initial Amount

Rearrange the formula to solve for \( N_0 \): \( N_0 = N(t) \times \left( 2 \right)^{\frac{t}{T_{1/2}}} \). Substitute the known values: \( N_0 = 1 \times 2^4 = 1 \times 16 = 16 \).
05

Solution Interpretation

This means that the initial concentration of plutonium-239 in the rock deposited 100,000 years ago was 16 mg/kg.

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Key Concepts

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

Half-life calculation
The concept of half-life is pivotal when studying radioactive decay. Simply put, it is the time required for half of a sample of a radioactive substance to decay. This process follows a predictable pattern, enabling us to calculate past or future amounts of the substance given its current amount and its half-life.The half-life formula is connected with exponential decay, described as:\[ N(t) = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]Here,
  • \( N(t) \) denotes the amount of substance left after time \( t \).
  • \( N_0 \) is the initial amount of substance.
  • \( T_{1/2} \) is the half-life period.
With this equation, whenever you know the current amount and the number of years elapsed, you can determine how much of the substance was initially present.
Plutonium-239
Plutonium-239 is a radioactive isotope known for its significance in both weaponry and nuclear power. Its long half-life of 25,000 years makes it persistent in the environment, meaning it decays slowly compared to many other isotopes. A fascinating aspect of Plutonium-239 is how its long half-life allows scientists to make inferences about historical events. For instance, scientists can date rocks or archaeological finds by measuring the amount of remaining plutonium-239. This attribute is why plutonium-239 was at the core of the original exercise, focusing on calculating how much was initially present in a rock deposited thousands of years ago.
Exponential decay formula
Radioactive substances like plutonium-239 decay exponentially. This means that the amount reduces at a rate proportional to its current value, rather than a fixed amount per unit time.The exponential decay formula, used in many scientific applications, is expressed as:\[ N(t) = N_0 \times e^{-\lambda t} \]But for half-life calculations specifically, we use:\[ N(t) = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]In this formula:
  • \( \lambda \) represents the decay constant 鈥� a unique value for each substance.
  • The relationship \( e^{-\lambda t} \) smoothly describes how quantities decrease exponentially over time.
This exponential nature is why small changes in time have a significant effect on how much substance is left, illustrated by how quickly the initial amount of plutonium-239 decreased in the original exercise.

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Most popular questions from this chapter

(Chemical Connections 9C) Why is high-energy radiation exposure of water in the body dangerous to rapidly dividing cells?

Match the radioactive isotope with its proper use: (a) Cobalt- 60 (b) Thallium-201 (c) Tritium (d) Mercury-197 1\. Heart scan during exercise 2\. Measure water content of body 3\. Kidney scan 4\. Cancer therapy

Americium-241, which is used in some smoke detectors, has a half-life of 432 years and is an alpha emitter. What is the decay product of americium- 241 , and approximately what percentage of the original americium-241 will be still around after 1000 years?

What are the products of the fusion of hydrogen- 2 and hydrogen-3 nuclei?

The most abundant isotope of uranium, \(238 \mathrm{U}\), does not undergo fission. Instead, it captures a neutron and emits two \(\beta\) particles to make a fissionable isotope of plutonium, which can then be used as fuel in a nuclear reactor. Write an equation for the nuclear reaction and identify the product nucleus.
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