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The decay constant of plutonium-239, a waste product from nuclear reactors, is \(2.88 \times 10^{-5}\) year \(^{-1}\). What is the half-life of \({ }^{239} \mathrm{Pu}\) ?

Short Answer

Expert verified
The half-life of \({}^{239}\mathrm{Pu}\) is approximately 24,062 years.

Step by step solution

01

Understanding Decay Constant and Half-Life

The decay constant (\(\lambda\)) is the probability per unit time that a given nuclei will decay. The half-life (\(T_{1/2}\)) is the time required for half of the radioactive nuclei in a sample to decay. They are related by the formula \(T_{1/2} = \frac{\ln(2)}{\lambda}\). Here, \(\lambda = 2.88 \times 10^{-5}\, \text{year}^{-1}\).

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Key Concepts

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

Decay Constant
The decay constant, often represented by the symbol \( \lambda \), is a fundamental concept in nuclear chemistry. It describes the likelihood that a single atom will decay in a given time period. Essentially, it is the decay rate of a radioactive substance. The value of the decay constant is unique for each radioactive element and is expressed in inverse time units, often per year in macro-level calculations.

When calculating the decay constant, consider that not all atoms in a radioactive sample decay simultaneously. Instead, the decay constant helps predict the average behavior of the entire sample, as it represents the probability of decay for each atom. This constant is crucial for understanding how quickly or slowly a substance goes through radioactive decay.

In the context of Plutonium-239 ({ }^{239} ext{Pu}), a radioactive isotope, its decay constant is given as \(2.88 \times 10^{-5}\, \text{year}^{-1}\). This means that each plutonium atom has a very small chance of decaying each year, but over many atoms, this probability becomes significant.
Half-Life
Half-life, denoted as \(T_{1/2}\), is a key term used in understanding radioactive decay. It is the duration of time needed for half of the radioactive nuclei in a sample to decay. This makes it a crucial measure for understanding how long a radioactive substance remains active.

The half-life is inversely related to the decay constant. This relationship can be expressed through the formula: \[T_{1/2} = \frac{\ln(2)}{\lambda}\]

The constant \(\ln(2)\) is derived from the natural logarithm of 2, which is approximately 0.693. Given the decay constant for Plutonium-239 is \(2.88 \times 10^{-5}\, \text{year}^{-1}\), using the formula allows us to calculate its half-life and understand how long this substance remains active in the environment or in nuclear waste.
Nuclear Chemistry
Nuclear chemistry is a branch of chemistry that deals with radioactive elements, nuclear reactions, and the properties and behaviors of nuclei. It encompasses both the study of naturally occurring elements and synthetic, man-made isotopes.

One significant area of nuclear chemistry is the study of radioactive decay processes, where unstable isotopes, like Plutonium-239, lose energy by emitting radiation. This process turns them into different isotopes or elements. Nuclear chemists study these processes to understand ensure safe handling and storage of radioactive materials.

Applications of nuclear chemistry reach various fields, including medicine (for imaging and treatment), power generation (in nuclear reactors), and environmental science (to track and manage radioactive pollution). Understanding the principles of nuclear chemistry allows scientists and engineers to utilize radioactive substances safely and effectively.
Plutonium-239
Plutonium-239 ( { }^{239} ext{Pu} ) is a radioactive isotope of the element plutonium, primarily known for its role in nuclear reactors and nuclear weapons. It is one of the primary waste products from nuclear reactors, weighing heavily on the management of nuclear waste due to its long half-life and the challenges of safe storage.

Plutonium-239 is noteworthy because it undergoes alpha decay, losing two protons and two neutrons to transform into Uranium-235, another significant radioactive isotope. This process reflects its stability and long-lasting presence in the environment, raising concerns about its implications on security and environmental health.

Handling Plutonium-239 requires stringent safety protocols due to its potential hazards, including radiation exposure and toxicity. Understanding its decay process and properties is crucial for its management in nuclear science and industry.

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Most popular questions from this chapter

The age of an igneous rock that has solidified from magma can be found by analyzing the amount of \({ }^{40} \mathrm{~K}\) and 40 Ar. If the rock contains \(1.20 \mathrm{mmol}\) of \({ }^{40} \mathrm{~K}\) and \(0.95 \mathrm{mmol}\) of \({ }^{40} \mathrm{Ar}\), how long ago did the rock cool? The half-life of potassium- 40 is \(1.25 \times 10^{9}\) years. $$ { }_{19}^{40} \mathrm{~K} \longrightarrow{ }_{18}^{40} \mathrm{Ar}+{ }_{1}^{0} \mathrm{e} $$

Why does a given nucleus have less mass than the sum of its constituent protons and neutrons?

What particle is produced in each of the following decay reactions? (a) \({ }_{11}^{24} \mathrm{Na} \rightarrow{ }_{12}^{24} \mathrm{Mg}+?\) (b) \({ }_{60}^{135} \mathrm{Nd} \rightarrow{ }_{59}^{135} \mathrm{Pr}+?\) (c) \({ }_{78}^{170} \mathrm{Pt} \rightarrow \frac{166}{76} \mathrm{Os}+?\)

Calculate the mass defect for the formation of an oxygen- 16 nucleus in both grams and \(\mathrm{g} / \mathrm{mol}\), and calculate the binding energy in both \(\mathrm{MeV} /\) nucleon and \(\mathrm{kJ} / \mathrm{mol.}\) The mass of an \({ }^{16} \mathrm{O}\) atom is \(15.99491 \mathrm{u}\).

The radioactive isotope \({ }^{100} \mathrm{~T}_{c}\) decays to form the stable isotope \({ }^{100}\) Mo. (a) There are two possible pathways for this decay. Write balanced equations for both. (b) Only one of the pathways is observed. Calculate the energy released by both pathways, and explain why only one is observed. Relevant masses are: \({ }^{100} \mathrm{~T} c(99.907657 \mathrm{u}),{ }^{100} \mathrm{Mo}\) \((99.90748 \mathrm{u}), \mathrm{e}^{-}(0.0005486 \mathrm{u})\)

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