Problem 28 - College Physics

Question

(a) What is the decay constant of fluorine- 17 if its half-life is known to be \(66.0 \mathrm{~s} ?\) (b) How long will it take for the activity of a sample of \(17 \mathrm{~F}\) to decrease to \(80 \%\) of its initial value? (c) Repeat part (b), but instead determine the time to decrease to an additional \(20 \%\) to \(60 \%\) of its initial value. Does it take twice as long to decay to \(60 \%\) compared to \(80 \%\) of its initial activity? Explain.

Step-by-Step Solution

Answer

Decay constant is 0.010506 s⁻¹. Time for 80% is 21.83 s; for 60%, 51.91 s. It does not take exactly twice as long.

Step 1: Decay Constant Formula

The decay constant \( \lambda \) is related to the half-life \( t_{1/2} \) by the formula: \[ \lambda = \frac{\ln(2)}{t_{1/2}} \] We need to substitute the given half-life of fluorine-17, \( t_{1/2} = 66.0 \) seconds, into this formula to find the decay constant.

Step 2: Calculate Decay Constant

Substitute \( t_{1/2} = 66.0 \) seconds into the decay constant formula:\[ \lambda = \frac{\ln(2)}{66.0} \approx 0.010506 \text{ s}^{-1} \] Thus, the decay constant \( \lambda \) is approximately \( 0.010506 \text{ s}^{-1} \).

Step 3: Activity Reduction to 80%

To find the time taken for the activity to reduce to 80%, we use the decay formula: \[ A = A_0 e^{-\lambda t} \] For 80% of the initial activity: \( A = 0.8 A_0 \), substitute into the formula and solve for \( t \):\[ 0.8 = e^{-0.010506 t} \]Take logs to solve for \( t \):\[ t = -\frac{\ln(0.8)}{0.010506} \approx 21.8325 \text{ s} \]

Step 4: Activity Reduction to 60%

Using the same formula, solve for the time taken for activity to reduce to 60%:For 60%: \( A = 0.6 A_0 \), plug into the formula:\[ 0.6 = e^{-0.010506 t} \]Take logs to solve for \( t \):\[ t = -\frac{\ln(0.6)}{0.010506} \approx 51.909 \text{ s} \]

Step 5: Compare Time Durations

Compare the time taken to reach 80% versus 60%: - To reach 80%: Approximately 21.83 seconds. - To reach 60%: Approximately 51.91 seconds. The time to reach 60% is not exactly twice the time to reach 80%. Generally, exponential decay does not equate to linear or arithmetic progression, meaning the time difference is greater than double due to the logarithmic nature of the decay equation.

Key Concepts

Decay Constant

The decay constant, often represented by the Greek letter \( \lambda \), is a crucial parameter in radioactive decay. It quantifies the probability of a radioactive nucleus decaying per unit time. In mathematical terms, the decay constant is related to the half-life \( t_{1/2} \) by the formula:

• \( \lambda = \frac{\ln(2)}{t_{1/2}} \)

For fluorine-17, with a half-life of 66 seconds, the decay constant is calculated using this formula. Substituting the given half-life value, we get:

• \( \lambda = \frac{\ln(2)}{66} \approx 0.010506 \, \text{s}^{-1} \)

This result means that, at any given second, there is roughly a 0.010506 probability that a fluorine-17 atom will decay. Understanding the decay constant helps in predicting how quickly a radioactive substance will decay over time.

Half-life

Half-life is a term used to describe the time required for half of a sample of radioactive material to decay. It is a characteristic property of each radioactive isotope and varies from isotope to isotope.

• For fluorine-17, the half-life is 66 seconds.

Knowing the half-life helps scientists and engineers estimate how long a radioactive substance will remain active and assist in safety planning and material handling. With a half-life of 66 seconds, fluorine-17 has a relatively short duration before a significant portion of it decays. This property is essential for applications that might use fluorine-17, such as in medical imaging where rapid decay is sometimes beneficial.

Exponential Decay

Exponential decay describes the process by which quantities decrease at rates proportional to their current value. This principle is fundamental in radioactive decay, leading to the characteristic exponential form of the decay equation:

• \( A = A_0 e^{-\lambda t} \)

In this equation:

• \( A \) is the remaining activity.
• \( A_0 \) is the initial activity.
• \( \lambda \) is the decay constant.
• \( t \) is the time elapsed.

This formula shows why radioactive decay isn’t simply a linear reduction. For example, the calculation shows that for fluorine-17:

• To decrease to 80% of its activity, it takes about 21.83 seconds.
• To further decrease to 60%, it takes a total of approximately 51.91 seconds.

Thus, the reduction to 60% takes more time than simply doubling the 80% time, due to the logarithmic nature of exponential decay. This insight is crucial for accurately predicting the behavior of decaying substances over time.

Fluorine-17

Fluorine-17 is a radioactive isotope of fluorine, with specific properties that make it interesting for scientific study. It’s characterized by:

• A short half-life of just 66 seconds.
• Known for undergoing beta-plus decay or electron capture.

Because of its rapid decay, fluorine-17 must be handled carefully and used promptly in experimental setups or medical applications. Its short existence lends itself well to experiments requiring quick turnaround, such as in tracers used in dynamic processes. For students and scientists, fluorine-17 offers a clear example of fast-paced radioactive decay, which can be observed and analyzed within a comparatively short span of time.