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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 faster due to its shorter half-life. After three half-lives, approximately 0.125 mg remains from the initial 1.00 mg sample for both isotopes.

Step by step solution

01

Determine the half-life of Americium-241

Using the formula for half-life, we can calculate Americium-241's half-life as follows: \[t_{1/2}^{Am-241} = \frac{\ln{2}}{k^{Am-241}} = \frac{\ln{2}}{1.6 \times 10^{-3}\,\mathrm{yr}^{-1}}\] Now, we need to plug in the numbers and solve for \(t_{1/2}^{Am-241}\). \[t_{1/2}^{Am-241} \approx 433\,\mathrm{yr}\] So, the half-life of Americium-241 is approximately 433 years.
02

Determine the half-life of Iodine-125

Using the same formula for half-life, we can calculate Iodine-125's half-life: \[t_{1/2}^{I-125} = \frac{\ln{2}}{k^{I-125}} = \frac{\ln{2}}{0.011\,\mathrm{day}^{-1}}\] Now, we need to plug in the numbers and solve for \(t_{1/2}^{I-125}\). \[t_{1/2}^{I-125} \approx 63\,\mathrm{days}\] So, the half-life of Iodine-125 is approximately 63 days.
03

Determine which isotope decays faster

By comparing the half-lives of Americium-241 and Iodine-125, we can see that Iodine-125 has a shorter half-life (63 days) compared to Americium-241 (433 years). Therefore, Iodine-125 decays at a faster rate.
04

Calculate the remaining mass after three half-lives

For both isotopes, we will calculate the remaining amount of a 1.00 mg sample after three half-lives using the formula: \[Mass_{remaining} = Mass_{initial} \times \left(\frac{1}{2}\right)^n\] where \(Mass_{remaining}\) is the mass remaining after \(n\) half-lives, \(Mass_{initial}\) is the initial mass, and \(n\) is the number of half-lives. For Americium-241: \[Mass_{remaining}^{Am-241} = 1.00\,\mathrm{mg} \times \left(\frac{1}{2}\right)^3 = 1.00\,\mathrm{mg} \times \frac{1}{8}\] \[Mass_{remaining}^{Am-241} \approx 0.125\,\mathrm{mg}\] For Iodine-125: \[Mass_{remaining}^{I-125} = 1.00\,\mathrm{mg} \times \left(\frac{1}{2}\right)^3 = 1.00\,\mathrm{mg} \times \frac{1}{8}\] \[Mass_{remaining}^{I-125} \approx 0.125\,\mathrm{mg}\] So, after three half-lives, approximately 0.125 mg remains from the 1.00 mg initial sample for both Americium-241 and Iodine-125.

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Key Concepts

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

Half-Life Calculation
In the study of radioactive decay, half-life is a fundamental concept. It represents the time taken for half of a radioactive substance to decay. The formula for calculating half-life \(t_{1/2}\) is expressed as \(t_{1/2} = \frac{\ln{2}}{k}\), where \(k\) is the decay rate constant. For different substances, the half-life can vary greatly, depending on how quickly they undergo decay. Let's consider the half-life of Americium-241 and Iodine-125. Americium-241 has a rate constant \(k = 1.6 \times 10^{-3} \mathrm{yr}^{-1}\). Plugging the values into our formula gives us a half-life of approximately 433 years. In contrast, Iodine-125 has a rate constant \(k = 0.011 \mathrm{day}^{-1}\), resulting in a much shorter half-life of about 63 days. This stark difference shows why Iodine-125 decays at a faster rate, due to its shorter half-life.
Rate Constants
The rate constant \(k\) is a critical component in understanding how quickly a radioactive isotope decays. It is measured in units of reciprocal time, such as \(\mathrm{yr}^{-1}\) or \(\mathrm{day}^{-1}\), denoting the frequency with which a substance undergoes decay within a given time frame. The relationship between the rate constant and half-life is inversely proportional. A higher rate constant means a shorter half-life and a faster rate of decay. For instance, the rate constant for Iodine-125 is 0.011 \(\mathrm{day}^{-1}\), resulting in a decay rate that rapidly reduces the substance within days rather than years. Conversely, Americium-241, with its lower rate constant of 1.6 \times 10^{-3} \(\mathrm{yr}^{-1}\), indicates a slower rate of decay, taking centuries for significant reduction.
Isotopes
In chemistry and physics, isotopes are defined as atoms with the same number of protons but different numbers of neutrons. This results in variations in their atomic mass but not in their chemical behavior. Isotopes of an element are generally noted to have different stabilities, which affects their rates of radioactive decay. Some isotopes, such as Iodine-125, are used in medical settings due to their predictable decay patterns and relatively short half-lives, making them suitable for diagnostic purposes. Others, like Americium-241, with longer half-lives, are utilized in devices like smoke detectors, where steady, long-term presence is beneficial.
Americium-241
Americium-241 is a synthetic element primarily utilized in smoke detectors. It demonstrates a long half-life, around 433 years, due to its slow decay rate constant of \(1.6 \times 10^{-3} \mathrm{yr}^{-1}\). This slow decay is ideal for long-term applications, where stability over decades is crucial.It emits alpha particles during its decay, which ionizes the air within the smoke detector. This ionization process allows the alarm mechanism to detect smoke by recognizing changes in the electrical current and ensuring safety from fires. Americium-241’s low decay rate ensures the longevity of smoke detectors without needing frequent replacement.
Iodine-125
Iodine-125 is a radioactive isotope with important applications in the field of medicine, particularly in diagnostic tests. It has a half-life of about 63 days, owing to its relatively high rate constant \(k = 0.011 \mathrm{day}^{-1}\). This makes it ideal for short-term medical applications such as tracing and imaging.In tests for thyroid function, Iodine-125's radiation properties can be harnessed due to its emission of gamma rays. Its half-life allows for effective diagnosis while minimizing prolonged radiation exposure, ensuring patient safety. Its quick decay means medical practitioners can use Iodine-125 to monitor bodily functions with precise, real-time results over a manageable period of time.

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Most popular questions from this chapter

The following mechanism has been proposed for the gas-phase reaction of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) and chlorine: Step 1: \(\mathrm{Cl}_{2}(g) \underset{k_{1}}{\stackrel{k_{1}}{\rightleftharpoons}} 2 \mathrm{Cl}(g) \quad\) (fast) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g) \quad\) (slow) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CCl}_{4} \quad\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest? Explain your answer. (a) \(E_{a}=45 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{~kJ} / \mathrm{mol}\)

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

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

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\)

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