Problem 73 - Chemistry: An Atoms-Focused Approach

Question

Radioactive isotopes such as \(^{32} \mathrm{P}\) are used to follow biological processes. The following radioactivity data (in relative radioactivity values) were collected for a sample containing \(^{32} \mathrm{P}\) : $$\begin{array}{cc} \text { Time (days) } & \text { Relative Radioactivity } \\\\\hline 0 & 10.00 \\\\\hline 1 & 9.53 \\\\\hline 2 & 9.08 \\\\\hline 5 & 7.85 \\\\\hline 10 & 6.16 \\\\\hline 20 & 3.79 \\\\\hline\end{array}$$ a. Write the rate law for the decay of \(^{32} \mathrm{P}\). b. Determine the value of the rate constant. c. Determine the half-life of \(^{32} \mathrm{P}\). d. How many days does it take for \(99 \%\) of a sample of \(^{32} \mathrm{P}\) to decay?

Step-by-Step Solution

Answer

Answer: The half-life of \(^{32} \mathrm{P}\) is approximately \(14.42\) days, and it takes approximately \(95.81\) days for \(99 \%\) of a sample to decay.

Step 1: Determine the rate law for the decay of \(^{32} \mathrm{P}\)

In radioactive processes, the rate of decay is proportional to the amount of radioactive material present. The rate law for radioactive decay can be written as: $$R(t) = kN(t)$$

Step 2: Determine the value of the rate constant

We can calculate the decay constant \(k\) using the given data. We will use any two data points to determine the exponential decay model, but it's better to choose points with a small difference in time. We will use data points for days 0 and 1: $$N(0) = 10.00$$ $$N(1) = 9.53$$ Using the decay formula: $$N(t) = N_0 e^{-kt}$$ We get: $$9.53 = 10.00 e^{-k(1)}$$ Solve for \(k\): $$k = -\ln(\frac{9.53}{10.00}) \approx 0.04805$$ The value of the decay constant \(k\) is approximately \(0.04805\).

Step 3: Determine the half-life of \(^{32} \mathrm{P}\)

Half-life is the time it takes for half of the radioactive material to decay. We can use the decay formula: $$N(t) = N_0 e^{-kt}$$ When half of the sample remains, \(N(t) = 0.5 N_0\). Plugging this into the decay formula and solving for the half-life \(t_{1/2}\), we get: $$0.5 N_0 = N_0 e^{-kt_{1/2}}$$ $$t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{0.04805} \approx 14.42 \text{ days}$$ The half-life of \(^{32} \mathrm{P}\) is approximately \(14.42\) days.

Step 4: Determine the time it takes for \(99 \%\) of a sample of \(^{32} \mathrm{P}\) to decay

We need to find the time \(t\) when \(99 \%\) of the sample has decayed, which means only \(1 \%\) of the sample remains: $$N(t) = 0.01 N_0$$ Plug this value into the decay formula and solve for \(t\): $$0.01 N_0 = N_0 e^{-kt}$$ $$t = \frac{\ln(0.01)}{k} \approx \frac{-4.605}{0.04805} \approx 95.81 \text{ days}$$ It takes approximately \(95.81\) days for \(99 \%\) of a sample of \(^{32} \mathrm{P}\) to decay.

Key Concepts

Radioactive Isotopes

Radioactive isotopes are versions of elements with unstable nuclei that decay over time, emitting radiation in the process. These isotopes have the same number of protons but differ in the number of neutrons. This instability is due to the imbalance between protons and neutrons. As the isotopes decay, they transform into different elements or isotopes by emitting particles and energy.
The radioactive isotope phosphorus-32 ( ^{32}P ) is a popular isotope used in biological sciences to track and study cellular processes. Since its radioactive decay can be measured over time, scientists often employ it to analyze growth patterns and metabolic activities in organisms. Knowledge about radioactive isotopes and their behavior provides valuable insights across various fields including chemistry, biology, archaeology, and medicine.

Exponential Decay Model

The exponential decay model is a mathematical representation used to describe the process by which a quantity decreases over time at a rate proportional to its current value. In the context of radioactive decay, it represents how the number of radioactive atoms decreases over time. The model can be expressed using the formula:
\[ N(t) = N_0 e^{-kt} \]
Here,

• \(N(t)\) is the quantity of substance that still remains after time \(t\).
• \(N_0\) is the initial quantity of the substance.
• \(k\) is the decay constant, indicating the rate of decay.

The exponential nature of this model means that the process of decay never truly stops but continues to decrease by halves over consistent periods. Understanding this concept is crucial for making predictions about how long it will take for a significant portion of a radioactive substance to decay.

Half-Life Calculation

Half-life is a critical concept in understanding radioactive decay, as it specifies the time required for half of the radioactive atoms in a sample to decay. This constant provides a reliable measure to track the decay process of isotopes.
To calculate the half-life, we use the formula:
\[ t_{1/2} = \frac{\ln(2)}{k} \]
Here,

• \(t_{1/2}\) is the half-life.
• \(k\) is the decay constant calculated from the data.

The logarithmic relationship means that the half-life does not depend on the initial or final amount of the substance; instead, it is intrinsic to the substance itself. For ^{32}P, for example, we found its half-life to be approximately 14.42 days. Knowing the half-life is extremely useful for applications in radiometric dating, medical treatments, and nuclear power generation.