Problem 105 - Chemistry The Central Science

Question

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?

Step-by-Step Solution

Answer

(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 1: 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 (t½) : \[ t_{1/2} = \frac{0.693}{k} \] 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: \( t_{1/2} = \frac{0.693}{0.011\ \mathrm{day}^{-1}} \)

Step 2: 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.

Step 3: 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

Step 4: 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.

Key Concepts

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.