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Radioactive iodine-131, which has a half-life of 8.04 days, is used in the form of sodium iodide to treat cancer of the thyroid. If you begin with \(25.0 \mathrm{mg}\) of \(\mathrm{Na}^{131} \mathrm{I},\) what quantity of the material remains after 31 days?

Short Answer

Expert verified
Approximately 1.82 mg of iodine-131 remains after 31 days.

Step by step solution

01

Understanding Half-Life

The concept of half-life is crucial here. Half-life is the time it takes for half of the radioactive substance to decay. For iodine-131, the half-life is 8.04 days.
02

Determine Number of Half-Lives

Calculate how many half-lives (8.04 days each) fit into 31 days. This is calculated as \[ n = \frac{31}{8.04} \approx 3.86 \] Thus, iodine-131 undergoes approximately 3.86 half-lives in 31 days.
03

Calculate Remaining Quantity

The remaining quantity of the substance can be determined using the formula \[ \text{final amount} = \text{initial amount} \times \left(\frac{1}{2}\right)^n \] Substitute the values:\[ \text{final amount} = 25.0 \times \left(\frac{1}{2}\right)^{3.86} \approx 25.0 \times 0.0729 \approx 1.82 \text{ mg} \] So, about 1.82 mg remains after 31 days.

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Key Concepts

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

Radioactive Decay
Radioactive decay is a natural process where unstable atomic nuclei lose energy by emitting radiation. This decay can result in the transformation of one element into another. It occurs at a predictable rate expressed by what we call "half-life," which is the time required for half of the radioactive atoms in a sample to decay. This makes radioactive decay both measurable and predictable.
Understanding radioactive decay is crucial in fields such as medicine, archaeology, and nuclear chemistry. It helps in dating fossils, managing nuclear waste, and treating diseases such as cancer. During decay, different types of radiation, like alpha, beta, or gamma radiation, can be emitted. Each has distinct properties and levels of penetration, which determine their effects and uses.
Iodine-131
Iodine-131 (\(_{53}^{131}\text{I}\)) is a radioactive isotope of iodine, widely used in medical applications, particularly for the treatment of thyroid cancer and hyperthyroidism. This is because iodine is naturally absorbed by the thyroid gland, ensuring effective targeting when administered.
One of the notable characteristics of iodine-131 is its half-life of 8.04 days, which is quite short compared to other radioactive substances. This means that after about 8 days, only half of the original iodine-131 remains active. Its ability to deliver a radioactive dose allows it to target and destroy unhealthy thyroid tissue effectively.
  • This property makes iodine-131 invaluable in nuclear medicine for both diagnostic and therapeutic purposes.
  • Patients receiving iodine-131 treatments are usually monitored carefully to manage any risks associated, given the radiation exposure.
Nuclear Chemistry
Nuclear chemistry is a branch of chemistry focused on the reactions involving atomic nuclei. This involves studies on both the spontaneous transformations that occur in nuclear decay and the reactions facilitated in reactors and particle accelerators.
It is essential for understanding how elements change, how nuclear energy is produced, and the development of new materials and energy sources. Nuclear chemistry provides insights into the mechanisms of radioactive decay, helping us understand isotopes like iodine-131.
  • Applications of nuclear chemistry extend from medical treatments to energy production, such as in nuclear power plants.
  • Safety and environmental implications are also key considerations as it involves managing radiation risks effectively.
  • Advancements in this field contribute to the development of cleaner and more efficient energy solutions.

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Most popular questions from this chapter

A proposed mechanism for the reaction of \(\mathrm{NO}_{2}\) and \(\mathrm{CO}\) is Step 1 Slow, endothermic $$2 \mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g})$$ Step 2 \(\quad\) Fast, exothermic $$\mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})$$ Overall Reaction Exothermic $$\mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})$$ (a) Identify each of the following as a reactant, product, or intermediate: \(\mathrm{NO}_{2}(\mathrm{g}), \mathrm{CO}(\mathrm{g}), \mathrm{NO}_{3}(\mathrm{g}), \mathrm{CO}_{2}(\mathrm{g})\) \(\mathrm{NO}(\mathrm{g})\) (b) Draw a reaction coordinate diagram for this reaction. Indicate on this drawing the activation energy for each step and the overall reaction enthalpy.

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 \(64 \mathrm{h} ?\)

Data for the following reaction are given in the table. $$2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NOBr}(\mathrm{g})$$ $$\begin{array}{llll}\hline \text { Experiment } & \begin{array}{l}{[\mathrm{NO}]} \\\\(\mathrm{M})\end{array} & \begin{array}{l}{\left[\mathrm{Br}_{2}\right]} \\\\(\mathrm{M})\end{array} & \begin{array}{l}\text { Initial Rate } \\\\(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s})\end{array} \\\\\hline 1 & 1.0 \times 10^{-2} & 2.0 \times 10^{-2} & 2.4 \times 10^{-2} \\\2 & 4.0 \times 10^{-2} & 2.0 \times 10^{-2} & 0.384 \\\3 & 1.0 \times 10^{-2} & 5.0 \times 10^{-2} & 6.0 \times 10^{-2} \\\\\hline\end{array}$$ (a) What is the order of the reaction with respect to \([\mathrm{NO}] ?\) (b) What is the order with respect to \(\left[\mathrm{Br}_{2}\right] ?\) (c) What is the overall order of the reaction?

Data for the reaction $$\begin{aligned}\left[\mathrm{Mn}(\mathrm{CO})_{5}\left(\mathrm{CH}_{3}\mathrm{CN}\right)\right]^{+}+\mathrm{NC}_{5}\mathrm{H}_{5} \longrightarrow & \\\&\left[\mathrm{Mn}(\mathrm{CO})_{5}\left(\mathrm{NC}_{5}\mathrm{H}_{5}\right)\right]^{+}+\mathrm{CH}_{3} \mathrm{CN}\end{aligned}$$ are given in the table. Calculate \(E_{\mathrm{a}}\) from a plot of \(\ln k\) versus \(1 / T\) $$\begin{array}{ll}\hline T(\mathrm{K}) & k\left(\min ^{-1}\right) \\\\\hline 298 & 0.0409 \\\308 & 0.0818 \\\318 & 0.157 \\\\\hline\end{array}$$

After five half-life periods for a first-order reaction, what fraction of reactant remains?

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