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Chapter 14: Problem 105

Americium-241 is used in smoke detectors. It has a first order 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 halflives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a 1.00 -mg sample of each isotope remains after 3 half-lives? (d) How much of a 1.00 -mg sample of each isotope remains after 4 days?

Short Answer

Expert verified
(a) Half-lives: Am-241: \(t_{1/2} \approx 433\ \mathrm{yr}\), I-125: \(t_{1/2} \approx 63\ \mathrm{days}\). (b) I-125 decays at a faster rate (shorter half-life). (c) Remaining amounts after 3 half-lives: Am-241: \(0.125\ \mathrm{mg}\), I-125: \(0.125\ \mathrm{mg}\). (d) Remaining amounts after 4 days: Am-241: \(0.991\ \mathrm{mg}\), I-125: \(0.656\ \mathrm{mg}\).

Step by step solution

01

Find the half-lives of Am-241 and I-125

To find the half-life of a radioactive isotope, we can use the relationship between the first-order rate constant (k) and the half-life (t陆) : \[ t_{1/2} = \frac{0.693}{k} \] For Americium-241 (Am-241): \( k = 1.6 \times 10^{-3}\ \mathrm{yr}^{-1} \) For Iodine-125 (I-125): \( k = 0.011\ \mathrm{day}^{-1} \) Now we can calculate the half-lives: Half-life of Am-241: \( t_{1/2} = \frac{0.693}{1.6 \times 10^{-3}\ \mathrm{yr}^{-1}} \) Half-life of I-125: \( t_{1/2} = \frac{0.693}{0.011\ \mathrm{day}^{-1}} \)
02

Compare the decay rates of Am-241 and I-125

We can compare the decay rates by comparing their half-lives. A shorter half-life means a faster decay rate.
03

Calculate the remaining amounts of isotopes after 3 half-lives

Using the half-life and the initial mass (1 mg), we can calculate the remaining amounts of each isotope after 3 half-lives. For a first-order decay, the remaining amount after n half-lives can be calculated using the equation: \( remaining\ amount = initial\ amount \times (\frac{1}{2})^n \) where, remaining amount = remaining amount of the isotope after n half-lives initial amount = initial amount of the isotope (1 mg) n = the number of half-lives
04

Calculate the remaining amounts of isotopes after 4 days

For this step, we need to use the decay constant (k) and the initial mass (1 mg) to calculate the remaining amounts of each isotope after 4 days. The remaining amount of an isotope after a certain time can be calculated using the equation: \( remaining\ amount = initial\ amount \times e^{-kt} \) where, remaining amount = remaining amount of the isotope after time t initial amount = initial amount of the isotope (1 mg) k = first-order rate constant t = time of decay (4 days) Here, we need to convert the time in days for Am-241. 1 year = 365 days, so 4 days = \(\frac{4}{365}\) years Now, we can calculate the remaining amounts of Am-241 and I-125 after 4 days.

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Key Concepts

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

Half-life Calculation
The half-life of a radioactive substance is the time it takes for half of the material to decay. This is a constant value for each isotope and is crucial for understanding how long a substance remains active. To calculate the half-life ( t_{1/2} ), we use the formula: \[ t_{1/2} = \frac{0.693}{k} \] where \( k \) is the first-order rate constant.
Let's calculate it for Americium-241 (Am-241) and Iodine-125 (I-125):
  • For Am-241: With \( k = 1.6 \times 10^{-3}\ \mathrm{yr}^{-1}\), the half-life is \(\frac{0.693}{1.6 \times 10^{-3}} \approx 433 \) years.
  • For I-125: With \( k = 0.011\ \mathrm{day}^{-1}\), the half-life is \(\frac{0.693}{0.011} \approx 63 \) days.
Understanding half-life allows us to determine how much of a substance remains over time, which is particularly important in applications like medical tests and smoke detectors.
First-order Rate Constant
In radioactive decay, the rate at which a substance decays is characterized by its first-order rate constant, \( k \). This constant is essential for calculating how quickly a radioactive substance will diminish.
The rate constant \( k \) represents the probability per unit time that a given atom will decay. High values of \( k \) indicate rapid decay, while low values suggest slower decay.
For Americium-241, the rate constant is \(1.6 \times 10^{-3}\ \mathrm{yr}^{-1}\), showing a slower decay over years. In contrast, Iodine-125 has a rate constant of \(0.011\ \mathrm{day}^{-1}\), meaning it decays significantly faster, reflecting its short-term medical use in tests.
Isotope Decay Comparison
When comparing isotopes, their decay rates and half-lives provide critical information about how quickly they diminish. By looking at both factors, it's easy to see which isotope will lose its radioactivity faster.
Here's how they compare:
  • Americium-241 (Am-241): With a half-life of around 433 years, it decays very slowly, making it suitable for long-term applications like smoke detectors.
  • Iodine-125 (I-125): With a half-life of about 63 days, it decays much faster, fitting its role in short-term medical diagnostics.
Shorter half-lives indicate quicker decay, which is why Iodine-125 decays faster than Americium-241. Understanding these differences helps in selecting the right isotope for specific purposes, ensuring effectiveness and safety in their applications.

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Most popular questions from this chapter

NO catalyzes the decomposition of \(\mathrm{N}_{2} \mathrm{O},\) possibly by the following mechanism: $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) & \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) & \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Why is NO considered a catalyst and not an intermediate? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism? If you think not, suggest what might be going on.

Enzymes are often described as following the two-step mechanism: $$ \begin{array}{l} \mathrm{E}+\mathrm{S} \rightleftharpoons \mathrm{ES} \quad(\text { fast }) \\ \mathrm{ES} \longrightarrow \mathrm{E}+\mathrm{P} \quad(\text { slow }) \end{array} $$ where \(\mathrm{E}=\) enzyme, \(\mathrm{S}=\) substrate, \(\mathrm{ES}=\) enzyme- substrate complex, and \(\mathrm{P}=\) product. (a) If an enzyme follows this mechanism, what rate law is expected for the reaction? (b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of \(\mathrm{E}\) with \(\mathrm{I}\), an inhibitor.

Molecular iodine, \(\mathrm{I}_{2}(g)\), dissociates into iodine atoms at \(625 \mathrm{~K}\) with a first-order rate constant of \(0.271 \mathrm{~s}^{-1}\). (a) What is the half-life for this reaction? (b) If you start with \(0.050 \mathrm{M} \mathrm{I}_{2}\) at this temperature, how much will remain after 5.12 s assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2} ?\)

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$ \begin{aligned} \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) \end{aligned} $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate constant, \(k ?\) (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M}\) ? (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C}\) ?

Consider two reactions. Reaction (1) has a constant half-life, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?
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