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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 half-lives 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) The half-lives of Americium-241 and Iodine-125 can be calculated as \(t_{1/2(Am-241)} = 433.5 \, \mathrm{yr}\) and \(t_{1/2(I-125)} = 63.1 \, \mathrm{days}\), respectively. b) Iodine-125 decays at a faster rate since its rate constant (\(0.011 \mathrm{day}^{-1}\)) is higher than that of Americium-241 (\(1.6 \times 10^{-3} \mathrm{yr}^{-1}\)). c) After 3 half-lives, the remaining amount of both Americium-241 and Iodine-125 will be \(0.125 \, \mathrm{mg}\) each. d) After 4 days, the remaining amount of Americium-241 is approximately \(0.994 \, \mathrm{mg}\), and the remaining amount of Iodine-125 is approximately \(0.641 \, \mathrm{mg}\).

Step by step solution

01

Half-life Formula for First-Order Decay Process

The half-life formula for a first-order decay process is given by: \(t_{1/2} = \frac{\ln{2}}{k}\), where \(t_{1/2}\) is the half-life and \(k\) is the first-order rate constant.
02

(a) Calculate half-lives of the isotopes

Using the given rate constant values, we will calculate the half-lives of both isotopes. For Americium-241 (\(k = 1.6 \times 10^{-3} \mathrm{yr}^{-1}\)): \(t_{1/2(Am-241)} = \frac{\ln{2}}{1.6 \times 10^{-3} \mathrm{yr}^{-1}}\) For Iodine-125 (\(k = 0.011 \mathrm{day}^{-1}\)): \(t_{1/2(I-125)} = \frac{\ln{2}}{0.011 \mathrm{day}^{-1}}\) Calculate these to find the half-lives.
03

(b) Determine which isotope decays faster

Compare the rate constants of the two isotopes to determine which one decays at a faster rate. The isotope with the higher rate constant decays faster.
04

(c) Calculate remaining amount after 3 half-lives

To calculate the remaining amount of each isotope after 3 half-lives, we will apply the following formula: Remaining amount = Initial amount × \((\frac{1}{2})^n\), where n = number of half-lives. For Americium-241: Remaining amount = 1.00 mg × \((\frac{1}{2})^3\) For Iodine-125: Remaining amount = 1.00 mg × \((\frac{1}{2})^3\) Calculate the remaining amount for both isotopes after 3 half-lives.
05

(d) Calculate remaining amount after 4 days

To calculate the remaining amount of each isotope after 4 days, we will use the formula: Remaining amount = Initial amount × \(e^{-kt}\), where t = time elapsed. For Americium-241: Remaining amount = 1.00 mg × \(e^{-(1.6 \times 10^{-3} \mathrm{yr}^{-1})(\frac{4 \, \mathrm{days}}{365.25 \, \mathrm{days/yr}})}\) For Iodine-125: Remaining amount = 1.00 mg × \(e^{-(0.011 \mathrm{day}^{-1})(4 \, \mathrm{days})}\) Calculate the remaining amount for both isotopes after 4 days.

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Key Concepts

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

First-Order Kinetics
In the world of chemistry, first-order kinetics is a fundamental concept, especially when discussing radioactive decay. A first-order decay process is characterized by the rate of change in the number of undecayed nuclei being proportional to the number of undecayed nuclei present. This means the decay process depends only on the amount present, not on any initial conditions or the passage of time.
For first-order reactions, one fascinating consequence is that they follow an exponential decay model. The mathematical representation is given as:
  • \[ \frac{dN}{dt} = -kN \ \] where \( N \ \) is the number of undecayed nuclei, \( \frac{dN}{dt} \ \) is the rate of decay, and \( k \ \) is the rate constant.
This formula helps us understand how substances decrease over time, providing the backbone for calculating half-lives and understanding radioactive processes.
Half-Life Calculation
The half-life of an isotope is a crucial concept in understanding radioactive decay. It's the time required for half of the radioactive nuclei in a sample to decay. This concept allows scientists to gauge how stable and long-lasting a radioactive substance might be.
For first-order kinetics, the half-life is determined using the formula:
  • \[ t_{1/2} = \frac{\ln{2}}{k} \]
Where \( t_{1/2} \ \) is the half-life and \( k \ \) is the first-order rate constant. This equation illustrates that the half-life is inversely related to the rate constant, meaning a higher rate constant results in a shorter half-life.
So for Americium-241 with a rate constant of \(1.6 \times 10^{-3} \mathrm{yr}^{-1}\), its half-life will be different compared to Iodine-125's \(0.011 \mathrm{day}^{-1}\). Calculating the half-life provides insights into how quickly each isotope reduces to half its original quantity.
Isotope Comparison
When comparing isotopes, understanding their rate constants and half-lives reveals significant characteristics about how they behave. Two isotopes with different rate constants will decay at different speeds, impacting their practical uses.
Americium-241, used in smoke detectors, is relatively stable with a smaller rate constant compared to Iodine-125. This means it has a longer half-life, ideal for consistent long-term performance in devices like smoke detectors. In contrast, Iodine-125 decays much faster, making it suitable for medical applications where rapid decay is beneficial.
So when comparing these isotopes:
  • Americium-241 has a longer half-life, making it practical for long-term use.
  • Iodine-125, with a shorter half-life, is apt for quick tests in medical fields.
This comparison showcases how understanding decay kinetics aids in selecting the appropriate isotope for specific applications.
Americium-241
Americium-241 is a synthetic radioactive isotope notable for its use in commercial smoke detectors. It undergoes alpha decay and emits ionizing radiation, which interacts with the smoke particles that enter the smoke detector. The rate constant for americium-241 is given as \(1.6 \times 10^{-3} \mathrm{yr}^{-1}\), suggesting a slow decay process.
This slow degradation is advantageous because it ensures the longevity of the smoke detector without frequent replacements. Americium-241's long half-life aligns well with this usage, providing reliability over years of use.
In summary, the characteristics of americium-241, such as its rate constant and half-life, make it uniquely suited for applications requiring stability and longevity.
Iodine-125
Iodine-125 is a radioactive isotope used primarily in medical diagnostics, particularly for thyroid function tests. With a rate constant of \(0.011 \mathrm{day}^{-1}\), it decays relatively quickly, making it ideal for applications that require rapid decay and short-term use.
The half-life of iodine-125, derived from its rate constant, means it maintains usefulness over a shorter period, which is perfect for medical conditions needing timely diagnostic results.
In medical tests, such as those for the thyroid, it effectively provides feedback on organ performance without exposing patients to long-lasting radiation. This quick action, thanks to its high rate constant, aligns with the need for fast results in healthcare environments.

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Most popular questions from this chapter

As described in Exercise 14.41, the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at \(660 \mathrm{~K}\) is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). (a) If we begin with an initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) pressure of 450 torr, what is the partial pressure of this substance after \(60 \mathrm{~s}\) ? (b) At what time will the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

The activation energy of an uncatalyzed reaction is \(95 \mathrm{~kJ} / \mathrm{mol}\). The addition of a catalyst lowers the activation energy to \(55 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C}\), (b) \(125^{\circ} \mathrm{C}\) ?

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the \(\mathrm{CO}_{2}\) accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

Cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) reacts with itself to form dicyclopentadiene $\left(\mathrm{C}_{10} \mathrm{H}_{12}\right)\(. A \)0.0400 \mathrm{M}\( solution of \)\mathrm{C}_{5} \mathrm{H}_{6}$ was monitored as a function of time as the reaction $2 \mathrm{C}_{5} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12}$ proceeded. The following data were collected: $$ \begin{array}{cc} \hline \text { Time (s) } & {\left[\mathrm{C}_{5} \mathrm{H}_{6}\right](M)} \\\ \hline 0.0 & 0.0400 \\ 50.0 & 0.0300 \\ 100.0 & 0.0240 \\ 150.0 & 0.0200 \\ 200.0 & 0.0174 \\ \hline \end{array} $$ Plot \(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, $\ln \left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\( versus time, and \)1 /\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]$ versus time. (a) What is the order of the reaction? (b) What is the value of the rate constant?

The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Based on the units for \(k\), is the reaction first or second order in \(\mathrm{NO}_{2}\) ? (b) If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M}\), how would you determine how long it would take for the concentration to decrease to \(0.025 \mathrm{M}\) ?

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