Problem 52 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A sample of \({ }_{92}^{233} \mathrm{U} \quad\left(T_{1}=1.59 \times 10^{5} \mathrm{yr}\right)\) contains \(5.50 \times 10^{18}\) nuclei. \(\quad(a)\) What \({ }^{2}\) is the decay constant? (b) Approximately how many disintegrations will occur per minute?

Step-by-Step Solution

Answer

The decay constant \( \lambda \) is approximately \(1.38 \times 10^{-12} \text{ s}^{-1}\) and around \(4.55 \times 10^{8}\) disintegrations occur per minute.

Step 1: Understand the Decay Constant Formula

To find the decay constant \( \lambda \), we use the formula \( \lambda = \frac{\ln(2)}{T_{1/2}} \), where \( T_{1/2} \) is the half-life of the substance. The natural logarithm of 2 is approximately \(0.693\).

Step 2: Calculate the Decay Constant

The half-life \( T_{1/2} \) is given as \(1.59 \times 10^{5}\) years. Convert this to seconds as we need the decay constant in per second: \( T_{1/2} = 1.59 \times 10^{5} \times 365 \times 24 \times 3600 \text{ seconds}\). Now, calculate \( \lambda \) using the formula: \( \lambda = \frac{0.693}{1.59 \times 10^{5} \times 365 \times 24 \times 3600} \approx 1.38 \times 10^{-12} \text{ s}^{-1}\).

Step 3: Find the Number of Disintegrations per Second

The number of disintegrations (activity) per second \( A \) is calculated using \( A = \lambda N \), where \( N \) is the number of nuclei, \( 5.50 \times 10^{18} \). So, \( A = (1.38 \times 10^{-12}) \times (5.50 \times 10^{18}) \approx 7.59 \times 10^{6} \text{ disintegrations per second} \).

Step 4: Convert Disintegrations per Second to Per Minute

To find the disintegrations per minute, multiply the disintegrations per second by 60: \( 7.59 \times 10^{6} \times 60 = 4.55 \times 10^{8} \text{ disintegrations per minute} \).

Key Concepts

Decay Constant

Radioactive decay constant, often symbolized as \( \lambda \), is a fundamental concept in understanding how unstable atoms lose energy over time. It describes the probability per unit time that a given nucleus will decay. Essentially, the decay constant tells us how "quickly" a radioactive substance undergoes decay.

The formula to calculate the decay constant is:

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

Here, \( \ln(2) \) is the natural logarithm of 2, approximately equal to 0.693. The variable \( T_{1/2} \) represents the half-life of the radioactive substance.

In practical applications, knowing the decay constant helps us predict how much of a radioactive substance remains after a certain period or how quickly it will decay. This is crucial in fields such as nuclear medicine and radiometric dating.

Half-life

The concept of half-life \( T_{1/2} \) in radioactive decay is a way to express how long it takes for half of the radioactive nuclei in a sample to decay. This time period is constant, meaning it doesn't change regardless of how much of the substance you have.

Imagine you have a lump of radioactive material. After one half-life, only half of the original radioactive nuclei will remain; the other half will have decayed into other elements. Another half-life later, half of what was left will decay, leaving a quarter of the original amount.

• This is why the process is exponential.
• The half-life provides a straightforward way of comparing the stability of different radioactive substances.

The half-life of Uranium-233, for example, is \(1.59 \times 10^{5}\) years. This means it takes that long for half of any given sample of Uranium-233 to decay.

Uranium-233

Uranium-233 is a radioactive isotope of uranium that is used in nuclear technology. It is of particular interest due to its fissile properties, which means it can sustain a nuclear chain reaction, making it useful in both nuclear power and weapons.

Derived from thorium-232, Uranium-233 has a relatively long half-life of \(1.59 \times 10^{5}\) years. This characteristic means it remains radioactive and useful as a nuclear fuel for extended periods. However, it also requires careful handling and storage to avoid environmental and health risks.

• Uranium-233 emits alpha particles during its decay process.
• These particles have significant energy and can damage living tissue if not properly contained.

Understanding Uranium-233's properties, including its decay constant and half-life, is essential for its effective and safe use in technology.