Problem 50 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Calculate the activity of a pure \(8.7-\mu \mathrm{g}\) sample of \({ }_{15}^{32} \mathrm{P}\left(T_{\frac{1}{2}}=1.23 \times 10^{6} \mathrm{~s}\right)\)

Step-by-Step Solution

Answer

The activity of the sample is calculated using the decay constant and the number of atoms.

Step 1: Identify the Given Values

We are given a phosphorus-32 (\( {}_{15}^{32}P \)) sample with a half-life \( T_{\frac{1}{2}} = 1.23 \times 10^6 \text{ s} \). The mass of the sample is \(8.7 \mu \text{g}\). We need to calculate the activity of this sample.

Step 2: Convert Mass to Moles

The atomic mass of phosphorus-32 is approximately 32 u, which corresponds to \(32 \times 1.66 \times 10^{-27} \text{ kg/mol} \). To find the number of moles in \(8.7 \mu \text{g}\), convert micrograms to grams (\(8.7 \mu \text{g} = 8.7 \times 10^{-6} \text{ g}\)) and then to moles: \[\text{Moles of } {}_{15}^{32}P = \frac{8.7 \times 10^{-6}}{32 \times 1.66 \times 10^{-27} \times 10^3} \text{ mol}\]

Step 3: Calculate the Number of Atoms

Use Avogadro's number \( (6.022 \times 10^{23} \text{ atoms/mol}) \) to calculate how many atoms are in the sample: \[\text{Number of atoms} = \text{Moles} \times 6.022 \times 10^{23}\]

Step 4: Calculate Decay Constant

The decay constant \( \lambda \) is related to the half-life by the formula: \[\lambda = \frac{0.693}{T_{1/2}}\] Substitute \( T_{1/2} = 1.23 \times 10^6 \text{ s} \) to find \( \lambda \).

Step 5: Calculate Activity

Activity \( A \) is given by the product of the decay constant and the number of radioactive atoms: \[A = \lambda \times N\] Substitute the values of \( \lambda \) and number of atoms \( N \) to calculate the activity in becquerels (\( \text{Bq} \)).

Key Concepts

Half-life Calculation

The concept of half-life is central to understanding radioactive decay. Half-life is the time it takes for half of a sample of a radioactive substance to decay. For instance, if you start with a certain amount of a radioactive element, after one half-life, only half of it will remain. This process continues until almost none remains.

Here are some key points to understand about half-life:

• Half-life is independent of the initial amount of the substance. Whether you have 1 gram or 100 grams, the time it takes for half to decay is the same.
• It is a statistical measure. Although individual atoms decay randomly, large groups of atoms display predictable behavior.
• In calculations, half-life helps determine the decay constant, which quantifies the rate of decay.

To calculate the decay constant from the half-life, use the formula: \[\lambda = \frac{0.693}{T_{1/2}}\]Here, \( T_{1/2} \) is the half-life, and \( \lambda \) is the decay constant. This formula reveals how the half-life inversely affects the rate at which the radioactive material decays.

Decay Constant

The decay constant, denoted as \( \lambda \), is a fundamental characteristic of radioactive decay. It represents the probability per unit time that a particular atom will decay.

Here is how the decay constant functions:

• The decay constant is directly related to the rate of decay. Higher decay constants mean a faster decay process.
• It is used in determining the radioactivity or activity of a substance, commonly measured in becquerels (Bq), where one Bq denotes one decay per second.
• The relationship between the decay constant and half-life is crucial for converting between the two, using the formula \( \lambda = \frac{0.693}{T_{1/2}} \).

In the original exercise, the decay constant of phosphorus-32 is calculated by plugging its half-life, \( 1.23 \times 10^6 \) seconds, into this formula. Understanding this constant allows for further calculations of how active a radioactive sample is or will be over time.

Phosphorus-32

Phosphorus-32 is a radioactive isotope of phosphorus, commonly used in various scientific applications due to its radioactive properties. It is an excellent example to illustrate radioactive decay concepts.

Details about phosphorus-32 include:

• Its half-life is relatively short, at approximately 14.3 days (or \( 1.23 \times 10^6 \) seconds), making it useful for experiments where short-lived radioactivity is needed.
• Phosphorus-32 emits beta particles during its decay process, which can be detected and measured easily, making observational research feasible.
• Used extensively in biochemical studies, particularly in labeling DNA or studies involving molecular biology due to its ability to integrate into biological molecules.

The exercise you're looking at includes phosphorus-32's decay characteristics, which involve converting mass into moles, determining the number of atoms through Avogadro's number, and finally calculating its activity using the decay constant obtained earlier. These interpretations of the isotope underline its practical significance and how its decay dynamics are computed and utilized.