Problem 7 - Applied Physics

Question

Find the source activity of a \(75.0-\mu\) Ci sample of \({ }_{84}^{214} \mathrm{Po}\) (polonium) \(5.00 \mu \mathrm{s}\) after certification. Its half- life is \(1.50 \times 10^{-4} \mathrm{~s}\).

Step-by-Step Solution

Answer

The source activity is approximately 76.74 µCi.

Step 1: Understand the decay formula

The decay of radioactive material can be described using the exponential decay formula: \( A = A_0 e^{-\lambda t} \), where \( A \) is the remaining activity, \( A_0 \) is the initial activity, \( \lambda \) is the decay constant, and \( t \) is the time elapsed.

Step 2: Calculate the decay constant

The decay constant \( \lambda \) is related to the half-life \( T_{1/2} \) of the substance by the formula \( \lambda = \frac{\ln(2)}{T_{1/2}} \). For polonium-214, \( T_{1/2} = 1.50 \times 10^{-4} \) seconds, so \( \lambda = \frac{\ln(2)}{1.50 \times 10^{-4}} \approx 4620 \text{ s}^{-1} \).

Step 3: Calculate the initial activity using the decay formula

We have \( A = 75.0 \mu \text{Ci} \) at \( t = 5.00 \mu\text{s} = 5.00 \times 10^{-6} \) seconds. Rearrange the decay formula to find \( A_0 \): \( A_0 = A e^{\lambda t} \). Substitute \( A = 75.0 \mu\text{Ci} \), \( \lambda \approx 4620 \text{ s}^{-1} \), and \( t = 5.00 \times 10^{-6} \text{ s} \).

Step 4: Calculate the source activity

Using \( A_0 = 75.0 \mu\text{Ci} \times e^{4620 \times 5.00 \times 10^{-6}} \), calculate \( A_0 \). This gives \( A_0 \approx 75.0 \mu\text{Ci} \times e^{0.0231} \approx 76.74 \mu\text{Ci} \).

Key Concepts

Exponential Decay Formula

Radioactive decay is best represented by the exponential decay formula, which models how the activity of a radioactive substance decreases over time. The equation is written as: \[ A = A_0 e^{-\lambda t} \]Where:

• \( A \) is the final activity observed at time \( t \),
• \( A_0 \) is the initial activity,
• \( \lambda \) is the decay constant, and
• \( t \) is the time that has elapsed.

The exponential nature of the formula means that the rate of decay is proportional to the current quantity. Therefore, as more of the radioactive substance decays, the speed of decay slows. This formula is essential for predicting how the activity of a material changes over time, allowing scientists and engineers to plan and manage the use of radioactive materials effectively.The exponential decay formula offers a practical way for understanding changes in any system that depletes in a non-linear manner, by providing all important parameters to solve for a known variable.

Decay Constant Calculation

The decay constant, represented by \( \lambda \), is a pivotal factor in understanding radioactive decay as it dictates how quickly a substance will decay. To calculate the decay constant, you can use the relationship it has with half-life.The formula is:\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]Here, \( T_{1/2} \) is the half-life of the substance. The half-life is the period it takes for half of the radioactive nuclei in a sample to decay. The natural logarithm of 2 is used because the decay process is exponential.When substituting the half-life for Polonium-214, which is \( 1.50 \times 10^{-4} \) seconds, we get:\[ \lambda \approx \frac{\ln(2)}{1.50 \times 10^{-4}} \approx 4620 \text{ s}^{-1} \]This decay constant is significant because it gives us an average rate at which the substance decays, providing important information for predicting the remaining activity after a given time period.

Half-Life

The concept of half-life is central to understanding radioactive decay. It is defined as the time required for half of the radioactive atoms in a sample to decay. For any given radioactive isotope, the half-life is a constant value, regardless of the initial amount or its current state of decay.Key points about half-life include:

• Each half-life results in the decay of 50% of the remaining radioactive atoms.
• The progress of decay is independent of initial activity size.
• Half-life enables the calculation of the remaining quantity of a substance at any point in time.

For example, Polonium-214 has a half-life of \( 1.50 \times 10^{-4} \) seconds. This means that within this very short period of time, half of any given quantity of Polonium-214 would decay. Understanding and using half-life in calculations enables precise monitoring and management of radioactive substances in various applications.

Polonium-214

Polonium-214 is an isotope of Polonium with some intriguing properties that are notable in the field of nuclear science. As part of the decay chain of Uranium-238, it exists only for a very short time due to its brief half-life.Key features of Polonium-214 include:

• It has a half-life of \( 1.50 \times 10^{-4} \) seconds, which is extremely short.
• It emits alpha particles during its decay process.
• It is part of a complex series of decays that eventually lead to a stable lead isotope.

The short half-life makes Polonium-214 significant mainly in studies related to nuclear processes and decay chains. The rapid rate of decay presents a challenge for handling this material outside of controlled environments.Understanding Polonium-214 is crucial for professionals dealing with radioactive decay, as its properties help in modeling decay processes and evaluating radioactive material safety.