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Americium-241 is used in smoke detectors. It has a rate constant for radioactive decay of \(k=1.6 \times 10^{-3} \mathrm{yr}^{-1}\). By contrast, iodine- 125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of \(k=0.011\) day \(^{-1} .\) (a) What are the half-lives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a \(1.00-\mathrm{mg}\) sample of either isotope remains after three half-lives?

Short Answer

Expert verified
The half-lives of Americium-241 and Iodine-125 are approximately 433 years and 63 days, respectively. Iodine-125 decays at a faster rate since it has a shorter half-life. After three half-lives, each 1.00 mg sample of either isotope (Americium-241 or Iodine-125) will have approximately 0.125 mg remaining.

Step by step solution

01

Calculate the half-lives of Americium-241 and Iodine-125

Use the decay constant formula (\(T_{1/2} = \frac{ln(2)}{k}\)), and the given rate constants for each isotope to calculate the half-lives. For Americium-241: \(T_{1/2(Am)} = \frac{ln(2)}{1.6 \times 10^{-3} \mathrm{yr}^{-1}}\) After solving, we get: \(T_{1/2(Am)} \approx 433 \mathrm{yr}\) For Iodine-125: \(T_{1/2(I)} = \frac{ln(2)}{0.011 \mathrm{day}^{-1}}\) After solving, we get: \(T_{1/2(I)} \approx 63 \mathrm{days}\)
02

Determine which isotope decays faster

Compare the half-lives of the two isotopes: The half-life of Americium-241 is approximately 433 years, while the half-life of Iodine-125 is approximately 63 days. Therefore, Iodine-125 decays at a faster rate since it has a shorter half-life.
03

Calculate the amount of the isotope remaining after three half-lives

Use the formula (\(N_t = N_0(1/2)^n\)) with n=3 for both isotopes. Here, the initial sample (\(N_0\)) is 1.00 mg. For Americium-241: \(N_t = 1.00 \times (1/2)^3\) After solving, we get: \(N_t \approx 0.125 \mathrm{mg}\) For Iodine-125: Similar to the calculation above, we obtain: \(N_t \approx 0.125 \mathrm{mg}\) After three half-lives, each 1.00 mg sample of either isotope (Americium-241 and Iodine-125) will have approximately 0.125 mg of the sample remaining.

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Key Concepts

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

Half-life Calculation
Understanding half-life is key to studying the stability of elements and the rate at which substances undergo radioactive decay. The half-life of a radioactive isotope is the time required for half the atoms in a given sample to decay. It's a constant value for each isotope, making it a handy tool for predicting the behavior of radioactive substances over time.

For calculating the half-life, we use the formula \( T_{1/2} = \frac{\ln(2)}{k} \), where \( T_{1/2} \) is the half-life and \( k \) is the decay constant. As shown in the textbook exercise, a lower decay constant means a longer half-life, suggesting that the substance takes more time to decay. When it comes to exercises involving half-life, it's important to understand that a substance with a shorter half-life decays more quickly, hence its greater activity. In comparing radioactive isotopes, this concept helps us determine which substance decays faster and anticipate the practical implications, such as safety guidelines for handling materials or the timing of diagnostic tests in nuclear medicine.
Decay Constant
The decay constant \( k \) represents the probability of decay of a nucleus per unit time. It's intrinsically linked to the stability of a nucleus; a high decay constant means a less stable nucleus and quicker decay. In the context of the textbook exercise, we can see that Americium-241 has a much lower decay constant than Iodine-125, which correlates to its longer half-life.

The decay constant is essential in various calculations in nuclear chemistry, not only for determining half-life but also in calculating the activity of a sample \( A = kN \) or the number of decays per unit time, where \( N \) is the number of undecayed nuclei. Knowing the decay constant enables us to estimate how quickly a radioactive sample will diminish to a safe level or how soon a radiopharmaceutical will lose its effectiveness. Thus, it holds significant importance in environmental studies, medical applications, and nuclear energy management.
Nuclear Chemistry
Nuclear chemistry delves into the reactions and properties of atomic nuclei. It covers various phenomena, including radioactive decay, nuclear fission, and fusion. Radioactive decay is a spontaneous process where unstable nuclei emit radiation to become more stable. There are different types of radioactive decay, such as alpha, beta, and gamma decay, each with unique implications for nuclear stability and the production of different isotopes.

In our textbook solution, we focus on the implications of decay in practical applications. Americium-241, used in smoke detectors, and Iodine-125, used in medical diagnostics, are both subjects of nuclear chemistry. Understanding decay rates and the effects of radiation are crucial not only for education but also for practical applications in medicine, where exact dosages of radioactive isotopes can aid in treatment or diagnostics, or in safety devices where longevity and reliability of the material are paramount.

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Most popular questions from this chapter

The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: \(\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-} .\) This rapid reaction gives the following rate data: \begin{tabular}{lll} \hline \(\left.\mathrm{OCl}^{-}\right](\mathrm{M})\) & {\(\left[\mathrm{I}^{-}\right](M)\)} & Rate \((M / \mathrm{s})\) \\ \hline \(1.5 \times 10^{-3}\) & \(1.5 \times 10^{-3}\) & \(1.36 \times 10^{-4}\) \\ \(3.0 \times 10^{-3}\) & \(1.5 \times 10^{-3}\) & \(2.72 \times 10^{-4}\) \\ \(1.5 \times 10^{-3}\) & \(3.0 \times 10^{-3}\) & \(2.72 \times 10^{-4}\) \\ \hline \end{tabular} (a) Write the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when \(\left[\mathrm{OCl}^{-}\right]=\) \(2.0 \times 10^{-3} M\) and \(\left[1^{-}\right]=5.0 \times 10^{-4} M\)

(a) What is meant by the term molecularity? (b) Why are termolecular elementary reactions so rare? (c) What is an intermediate in a mechanism?

(a) What part of the energy profile of a reaction is affected by a catalyst? (b) What is the difference between a homogeneous and a heterogeneous catalyst?

A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if (a) \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ;\) (b) \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?\)

The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: \begin{tabular}{ll} \hline Temperature (K) & Rate Constant (s \(^{-1}\) ) \\ \hline 300 & \(3.2 \times 10^{-11}\) \\ 320 & \(1.0 \times 10^{-9}\) \\ 340 & \(3.0 \times 10^{-8}\) \\ 355 & \(2.4 \times 10^{-7}\) \\ \hline \end{tabular} From these data, calculate the activation energy in units of \(\mathrm{kJ} / \mathrm{mol}\)

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