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

Short Answer

Expert verified
Approximately 3.1% of radioactive copper (II) acetate remains after 64 hours.

Step by step solution

01

Understanding Half-Life

The half-life of a substance is the time it takes for half of the radioactive substance to decay. For this isotope, the half-life is given as 12.70 hours, meaning every 12.70 hours, half of the initial amount of the substance will have decayed.
02

Determine Number of Half-Lives

To find out how many half-lives have passed in 64 hours, divide the total time by one half-life: \[ \text{Number of half-lives} = \frac{64}{12.7} \approx 5.04. \]
03

Calculate Remaining Fraction

The fraction of a radioactive substance remaining after a certain number of half-lives can be calculated using the formula \( \left(\frac{1}{2}\right)^n \), where \( n \) is the number of half-lives. Here, \( n \approx 5.04 \), so we compute: \[ \left(\frac{1}{2}\right)^{5.04} \approx 0.031. \] Therefore, approximately 3.1% of the radioactive copper (II) acetate 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 Calculation
Half-life calculation is a fundamental concept in understanding radioactive decay. It refers to the time required for half of a quantity of a radioactive substance to transform into a different element or isotope through decay processes.
When computing the half-life, you essentially calculate how long it takes for an initial sample to reduce to 50% of its original activity or mass. In mathematical terms, the remaining quantity of a substance after a certain number of half-lives can be found using the formula:
  • \( \text{Remaining Fraction} = \left(\frac{1}{2}\right)^n \)

Where \( n \) is the number of half-lives that have passed. For example, after one half-life, 50% remains; after two half-lives, 25% remains, and so on.
To determine the number of half-lives \( n \), you divide the total elapsed time by the duration of one half-life, as demonstrated in the exercise with the isotope \(^{64} \mathrm{Cu}\).
By understanding this calculation process, you can predict how much of a radioactive substance will be left after any given time, which is essential in fields like medicine, which utilize radioisotopes for research and treatment.
Isotope Decay
Isotope decay involves the transformation of an unstable isotope into a more stable atom through the emission of radiation. This process is natural and occurs in isotopes that have an unstable nucleus. Over time, as isotopes decay, they release energy in the form of particles or electromagnetic waves.
This can include:
  • Alpha decay - emission of alpha particles (Helium nuclei)
  • Beta decay - transformation of a neutron into a proton or vice versa
  • Gamma decay - emission of high-energy photons

The type of decay an isotope undergoes depends on its specific properties. For \(^{64} \mathrm{Cu}\), a radioactive isotope, decay results in the loss of particles and energy, changing the makeup of the isotope until it reaches a stable form.
Understanding decay is crucial in various scientific and industrial applications. It enables us to harness isotopes for medical diagnostics, treatment, and research, such as the study of Wilson's disease through copper isotopes.
Wilson's Disease Study
Wilson's Disease is a genetic disorder that affects the body's ability to manage copper levels efficiently, leading to excessive accumulation primarily in the liver and brain. This disease can cause severe neurological and hepatic damage if not managed properly.
Studying Wilson's Disease involves techniques such as using radioactive isotopes like \(^{64} \mathrm{Cu}\). Researchers utilize this specific copper isotope due to its properties and half-life, providing insights into how the disease affects copper metabolism within the body.
The use of copper isotopes helps:
  • Track the absorption and utilization of copper within the body
  • Explore effective treatments and management of the condition
  • Develop targeted therapies to reduce copper accumulation

Through such studies, scientists and medical professionals aim to further understand the pathophysiology of Wilson's Disease and improve patient outcomes by finding more effective strategies for managing copper levels.

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Most popular questions from this chapter

Experimental data are listed here for the reaction \(\mathrm{A} \longrightarrow 2 \mathrm{B}.\) $$\begin{array}{ll}\hline \begin{array}{l}\text { Time } \\\\(\mathrm{s})\end{array} & \begin{array}{l}{[\mathrm{B}]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} \\\\\hline 0.00 & 0.000 \\\10.0 & 0.326 \\\20.0 & 0.572 \\\30.0 & 0.750 \\\40.0 & 0.890 \\\\\hline\end{array}$$ (a) Prepare a graph from these data, connect the points with a smooth line, and calculate the rate of change of [B] for each \(10-\) s interval from 0.0 to 40.0 s. Does the rate of change decrease from one time interval to the next? Suggest a reason for this result. (b) How is the rate of change of \([\mathrm{A}]\) related to the rate of change of \([\mathrm{B}]\) in each time interval? Calculate the rate of change of \([\mathrm{A}]\) for the time interval from 10.0 to \(20.0 \mathrm{s}\) (c) What is the instantaneous rate when \([\mathrm{B}]=0.750\) \(\mathrm{mol} / \mathrm{L} ?\)

Ammonia decomposes when heated according to the equation $$\mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{NH}_{2}(\mathrm{g})+\mathrm{H}(\mathrm{g})$$ The data in the table for this reaction were collected at a high temperature. $$\begin{array}{ll}\hline \begin{array}{l}\text { Time } \\\\\text { (h) }\end{array} & \begin{array}{l}{\left[\mathrm{NH}_{3}\right]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} \\\\\hline 0 & 8.00 \times 10^{-7} \\\25 & 6.75 \times 10^{-7} \\\50 & 5.84 \times 10^{-7} \\\75 & 5.15 \times 10^{-7} \\\\\hline\end{array}$$ Plot \(\ln \left[\mathrm{NH}_{3}\right]\) versus time and \(1 /\left[\mathrm{NH}_{3}\right]\) versus time. What is the order of this reaction with respect to NH \(_{3} ?\) Find the rate constant for the reaction from the slope.

Describe each of the following statements as true or false. If false, rewrite the sentence to make it correct. (a) The rate-determining elementary step in a reaction is the slowest step in a mechanism. (b) It is possible to change the rate constant by changing the temperature. (c) As a reaction proceeds at constant temperature, the rate remains constant. (d) A reaction that is third order overall must involve more than one step.

The compound \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) decomposes in a first-order reaction to elemental Xe with a half-life of 30. min. If you place \(7.50 \mathrm{mg}\) of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) in a flask, how long must you wait until only 0.25 mg of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) remains?

Give the relative rates of disappearance of reactants and formation of products for each of the following reactions. (a) \(2 \mathrm{O}_{3}(\mathrm{g}) \longrightarrow 3 \mathrm{O}_{2}(\mathrm{g})\) (b) \(2 \mathrm{HOF}(\mathrm{g}) \longrightarrow 2 \mathrm{HF}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\)

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