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The radioactive isotope \(^{64} \mathrm{Cu}\) is used in the form of copper(II) acetate to study Wilson's disease. The isotope has a half-life of 12.70 hours. What fraction of radioactive copper(II) acetate remains after 64 hours?

Short Answer

Expert verified
Approximately 3.13% of radioactive copper(II) acetate remains after 64 hours.

Step by step solution

01

Understand the Half-Life Definition

The half-life of a radioactive substance is the time required for half of the substance to decay. For the isotope \(^{64} \mathrm{Cu}\), the half-life is given as 12.70 hours.
02

Calculate the Number of Half-Lives

To find the total number of half-lives that have passed in 64 hours, divide 64 hours by the half-life of 12.70 hours:\[\text{Number of half-lives} = \frac{64}{12.70} \approx 5.04\]
03

Use the Exponential Decay Formula

The fraction remaining after \( n \) half-lives can be calculated using the formula:\[\left(\frac{1}{2}\right)^n\]Where \( n = 5.04 \).
04

Calculate the Remaining Fraction

Substitute \( n = 5.04 \) into the decay formula:\[\left(\frac{1}{2}\right)^{5.04} \approx 0.0313\]This result represents the fraction of the radioactive copper(II) acetate that remains after 64 hours.

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Key Concepts

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

half-life
In radioactive decay, the concept of half-life is essential to understanding how substances change over time. The half-life is the period it takes for half of a sample of a radioactive substance to decay and lose its radioactivity. This means that after one half-life, only half of the original substance remains radioactive.
When calculating how much of a substance remains after a certain time, we use the idea of half-lives to break the total time into smaller intervals. In the context of the copper-64 exercise, the half-life was given as 12.70 hours. If we wish to know the remaining fraction after 64 hours, we first determine how many half-lives fit into this period. By dividing 64 by 12.70, we find approximately 5.04 half-lives have passed.
This number helps us understand that the radioactive isotope has significantly decayed over this time frame. Using the formula \( \left(\frac{1}{2}\right)^n \) where \( n \) is the number of half-lives, we quantify the fraction remaining. In our example, \( n = 5.04 \), thus resulting in a fraction of about 0.0313 as remaining from the original sample.
Wilson's disease
Wilson's disease is a rare genetic disorder that causes excessive copper accumulation in organs like the liver and brain. Our bodies typically regulate copper absorption and excretion, but for individuals with Wilson's disease, this regulation is impaired, leading to copper buildup. Over time, the excess copper causes damage to tissues and organs.
To study the disease and its effects, researchers often use tracers such as the radioactive isotope copper-64. These tracers help scientists observe how copper moves and where it accumulates in the body. This information is crucial in diagnosing the disease and determining the appropriate treatment measures.
By understanding how copper-64 behaves in the body, doctors can make more informed decisions about managing the disease, such as assessing how much copper is deposited in specific organs. Early detection and treatment of Wilson's disease are key to preventing long-term complications.
copper-64
Copper-64, or \(^{64}\text{Cu}\), is a radioactive isotope utilized in medical research and diagnostics because of its unique properties. Unlike stable copper, copper-64 emits radiation, which can be detected and tracked by special imaging equipment. This feature makes it especially valuable in studying diseases that involve copper metabolism, like Wilson's disease.
One of the primary attributes of copper-64 is its half-life of 12.70 hours. This relatively short half-life is advantageous because it limits the duration of radiation exposure to patients. Despite its short-lived nature, copper-64 provides enough time for thorough analysis and imaging in medical procedures.
Copper-64 is typically incorporated into compounds like copper(II) acetate, allowing it to be absorbed and utilized in biological studies. By observing copper-64's path through the body, researchers gain valuable insights into how copper metabolism is altered in certain conditions, thereby aiding in the treatment and study of related disorders.

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Most popular questions from this chapter

Radioactive gold-198 is used in the diagnosis of liver problems. The half-life of this isotope is 2.7 days. If you begin with a 5.6-mg sample of the isotope, how much of this sample remains after 1.0 day?

Nitrosyl bromide, NOBr, is formed from \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\) : $$2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{g})$$ Experiments show that this reaction is second order in NO and first order in \(\mathrm{Br}_{2}\). (a) Write the rate equation for the reaction. (b) How does the initial reaction rate change if the concentration of \(\mathrm{Br}_{2}\) is changed from \(0.0022 \mathrm{mol} / \mathrm{L}\) to \(0.0066 \mathrm{mol} / \mathrm{L} ?\) (c) What is the change in the initial rate if the concentration of NO is changed from \(0.0024 \mathrm{mol} / \mathrm{L}\) to \(0.0012 \mathrm{mol} / \mathrm{L} ?\)

Identify which of the following statements are incorrect. If the statement is incorrect, rewrite it to be correct. (a) Reactions are faster at a higher temperature because activation energies are lower. (b) Rates increase with increasing concentration of reactants because there are more collisions between reactant molecules. (c) At higher temperatures, a larger fraction of molecules have enough energy to get over the activation energy barrier. (d) Catalyzed and uncatalyzed reactions have identical mechanisms.

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}),\) decomposes to \(\mathrm{H}_{2} \mathrm{O}(\ell)\) and \(\mathrm{O}_{2}(\mathrm{g})\) in a reaction that is first order in \(\mathrm{H}_{2} \mathrm{O}_{2}\) and has a rate constant \(k=1.06 \times 10^{-3} \mathrm{min}^{-1}\) at a given temperature. (a) How long will it take for \(15 \%\) of a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) to decompose? (b) How long will it take for \(85 \%\) of the sample to decompose?

The decomposition of dinitrogen pentaoxide $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ has the following rate equation: Rate \(=k\left[\mathrm{N}_{2} \mathrm{O}_{5}\right] .\) It has been found experimentally that the decomposition is \(20.5 \%\) complete in 13.0 hours at \(298 \mathrm{K}\). Calculate the rate constant and the half-life at \(298 \mathrm{K}.\)

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