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Define "third-life" in a similar way to "half-life," and determine the "third- life" for a nuclide that has a half-life of 31.4 years.

Short Answer

Expert verified
The third-life for a nuclide, defined similarly to half-life as the time taken for a substance to reduce to one-third of its initial amount, can be calculated using the relationship \( T_{1/3} = \frac{\ln{\frac{1}{3}}}{\ln{\frac{1}{2}}} \times T_{1/2} \). For a nuclide with a half-life of 31.4 years, its third-life is approximately 20.0 years.

Step by step solution

01

Defining Half-life and Third-life

Half-life is defined as the time taken for a substance to reduce to half its initial amount. Similarly, we can define "third-life" as the time taken for a substance to reduce to one-third of its initial amount.
02

Relationship between Third-life and Half-life

Let's consider a substance that has a half-life of T years. After this time, the remaining amount of the substance will be half of the initial amount. To obtain the third-life, we need to find the time it takes for the substance to reduce to one third of its initial amount. To do this, let's look at the general equation that governs this kind of decay: \( N_t = N_0 \times \frac{1}{2}^{\frac{t}{T_{1/2}}} \) Where: - \(N_t\) is the remaining amount of the substance after t years. - \(N_0\) is the initial amount of the substance. - \(T_{1/2}\) is the half-life of the substance. To determine the third-life (\(T_{1/3}\)), we need to modify the equation so that it represents the remaining amount of the substance equal to one-third of the initial amount. Therefore, the equation becomes: \( N_t = N_0 \times \frac{1}{3}^{\frac{t}{T_{1/3}}} \) Now, we have to find a relationship between \(T_{1/2}\) and \(T_{1/3}\) to solve the problem.
03

Deriving the Relationship between Third-life and Half-life

First, let's find the ratio of the initial amount (\(N_0\)) to the remaining amount (\(N_t\)) at the half-life: \( \frac{N_0}{N_t} = \frac{1}{2} \) Similarly, let's find the ratio of the initial amount (\(N_0\)) to the remaining amount (\(N_t\)) at the third-life: \( \frac{N_0}{N_t} = \frac{1}{3} \) Now let's equate both expressions in terms of the half-life and third-life: \( \frac{1}{2}^{\frac{t}{T_{1/2}}} = \frac{1}{3}^{\frac{t}{T_{1/3}}} \) To achieve a relationship between \(T_{1/2}\) and \(T_{1/3}\), let's take the natural logarithm of both sides: \( \frac{t}{T_{1/2}} \times \ln{\frac{1}{2}} = \frac{t}{T_{1/3}} \times \ln{\frac{1}{3}} \) Divide both sides by t: \( \frac{\ln{\frac{1}{2}}}{T_{1/2}} = \frac{\ln{\frac{1}{3}}}{T_{1/3}} \) Now, we can obtain the relationship: \( T_{1/3} = \frac{\ln{\frac{1}{3}}}{\ln{\frac{1}{2}}} \times T_{1/2} \)
04

Calculating Third-life

Using the given half-life of the nuclide, T = 31.4 years, let's find the third-life (T_{1/3}): \( T_{1/3} = \frac{\ln{\frac{1}{3}}}{\ln{\frac{1}{2}}} \times 31.4 \approx 20.0 \) years So, the third-life for the nuclide with a half-life of 31.4 years is approximately 20.0 years.

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Key Concepts

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

Third-life
In the world of radioactive decay, we frequently hear about half-life—the time it takes for half of a radioactive substance to decay. However, there's another less commonly discussed concept: third-life. Third-life is the time duration required for a radioactive substance to reduce to one-third of its initial amount. It's an alternative way to measure radioactive decay rates and offers a broader understanding of this process.

To calculate the third-life, you use a similar approach to half-life calculations. By using specific equations related to decay, you can determine how long it will take for only one-third of the original nuclide to remain. This concept, like half-life, is critical in fields such as nuclear physics and radioactive dating, as it can help estimate timelines concerning nuclear reactions and material dating.
Radioactive Decay
Radioactive decay is a natural process where unstable atomic nuclei lose energy by emitting radiation. This decay process allows the nuclei to transform into a new state, often forming different elements or isotopes.

There are several types of radioactive decay, such as alpha decay, beta decay, and gamma decay. Each type involves the emission of different particles or electromagnetic radiation, contributing to the transformation of a nuclide.
  • Alpha decay emits alpha particles, consisting of two protons and two neutrons.
  • Beta decay involves the conversion of neutrons into protons with the release of electrons or positrons.
  • Gamma decay emits high-energy photons without changing the number of protons or neutrons.

This decay is essential in various applications, from carbon dating archaeological artifacts to medical treatments and power generation in nuclear reactors. Understanding it is crucial for any applications involving radioactive materials.
Nuclides
Nuclides are distinct species of atoms characterized by the number of protons and neutrons within their nuclei. Each nuclide corresponds to a unique atomic configuration and has specific properties, such as mass and stability.

Every element on the periodic table can have multiple nuclides, known as isotopes, differing only in the number of neutrons. These isotopes may be stable or radioactive, the latter undergoing decay to reach a more stable state. Some nuclides are used in industry and medicine due to their radioactive nature.

The consistency in atomic and mass numbers helps in identifying nuclides, making it possible to predict their behavior and energy levels during radioactive decay. This understanding assists physicists and chemists in various applications such as tracing chemical pathways or understanding stellar processes.
Exponential Decay
Exponential decay describes how the quantity of something decreases at a rate proportional to its current value. In the context of radioactive decay, exponential decay models the process through which a radioactive substance diminishes over time.

The mathematical representation for this process is given by:\[ N(t) = N_0 \times e^{-\lambda t} \]Where:
  • \( N(t) \) is the quantity at time \( t \).
  • \( N_0 \) is the initial quantity.
  • \( \lambda \) is the decay constant.

Exponential decay is a universal concept appearing also in areas like finance and biology. Its characteristic feature is a steady ratio of decrease, making predictions and calculations straightforward once the decay constant is known. Observing how substances decline exponentially helps develop more accurate models for predicting changes over time, critical in fields utilizing radioactive materials.
Natural Logarithm
The natural logarithm is a mathematical function denoted as \( \ln(x) \). It is the logarithm to the base \( e \), where \( e \) (approximately 2.718) is the constant of natural growth.

In the context of nuclear physics and radioactive decay, natural logarithms are crucial for calculating decay over time, such as half-life and third-life calculations. They transform multiplicative relationships into additive ones, simplifying the process of determining decay timelines.

For instance, when determining the time for half or one-third of a substance to remain, taking the natural logarithm allows you to rearrange and solve exponential decay equations easily. The property of turning multiplication into addition (through logarithms) greatly facilitates complex calculations involving rates and exponential functions, making them indispensable in sciences involving continuous growth or decay.

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Most popular questions from this chapter

One type of commercial smoke detector contains a minute amount of radioactive americium-241 ( \(\left.^{(241} \mathrm{Am}\right),\) which decays by \(\alpha\) -particle production. The \(\alpha\) particles ionize molecules in the air, allowing it to conduct an electric current. When smoke particles enter, the conductivity of the air is changed and the alarm buzzes. a. Write the equation for the decay of \(^{241}_{95} \mathrm{Am}\) by \(\alpha\) -particle production. b. The complete decay of \(^{241} \mathrm{Am}\) involves successively \(\alpha, \alpha\) \(\boldsymbol{\beta}, \alpha, \alpha, \boldsymbol{\beta}, \alpha, \alpha, \alpha, \boldsymbol{\beta}, \alpha,\) and \(\boldsymbol{\beta}\) production. What is the final stable nucleus produced in this decay series? c. Identify the 11 intermediate nuclides.

The only stable isotope of fluorine is fluorine-19. Predict possible modes of decay for fluorine-21, fluorine-18, and fluorine-17.

The most significant source of natural radiation is radon-222. \(^{222} \mathrm{Rn},\) a decay product of \(^{238} \mathrm{U},\) is continuously generated in the earth's crust, allowing gaseous Rn to seep into the basements of buildings. Because \(^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of 3.82 days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \(^{238} \mathrm{U}\) decays to \(^{222} \mathrm{Rn} ?\) What nuclei are produced when \(^{222} \mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \(^{222} \mathrm{Rn} ?\) c. Another problem associated with \(^{222} \mathrm{Rn}\) is that the decay of \(^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer \(\left(t_{1 / 2}=\right.\) 3.11 min) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by \(\alpha\) particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that \(^{222}\) Rn levels not exceed 4 pCi per liter of air (1 \(\mathrm{Ci}=1\) curie \(=3.7 \times 10^{10}\) decay events per second; \(1 \mathrm{pCi}=1 \times 10^{-12} \mathrm{Ci}\). Convert \(4.0 \mathrm{pCi}\) per liter of air into concentrations units of \(^{222} \mathrm{Rn}\) atoms per liter of air and moles of \(^{222}\) Rn per liter of air.

A rock contains \(0.688 \mathrm{mg}^{206} \mathrm{Pb}\) for every \(1.000 \mathrm{mg}\) \(^{238} \mathrm{U}\) present Assuming that no lead was originally present, that all the \(^{206} \mathrm{Pb}\) formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between \(^{238} \mathrm{U}\) and \(^{206} \mathrm{Pb}\) is negligible, calculate the age of the \(\text { rock. For }^{238} \mathbf{U}, t_{1 / 2}=4.5 \times 10^{9} \text { years. }\)

Naturally occurring uranium is composed mostly of \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U},\) with relative abundances of \(99.28 \%\) and \(0.72 \%,\) respectively. The half-life for \(^{238} \mathrm{U}\) is \(4.5 \times 10^{9}\) years, and the half-life for \(^{235} \mathrm{U}\) is \(7.1 \times 10^{8}\) years. Assuming that the earth was formed 4.5 billion years ago, calculate the relative abundances of the \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U}\) isotopes when the earth was formed.

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