Problem 19 - College Physics

Question

A sample of the radioactive nuclide \(^{199} \mathrm{Pt}\) is prepared that has an initial activity of \(7.56 \times 10^{11}\) Bq. (a) 92.4 min after the sample is prepared, the activity has fallen to \(9.45 \times 10^{10}\) Bq. What is the half-life of this nuclide? (b) How many radioactive nuclei were initially present in the sample?

Step-by-Step Solution

Answer

The half-life of the nuclide is approximately 25.7 minutes, and there were initially about \(2.8 \times 10^{13}\) radioactive nuclei present.

Step 1: Understand the Decay Formula

The radioactive decay formula is given by:\[A = A_0 e^{-\lambda t}\]where:- \(A\) is the activity at time \(t\),- \(A_0\) is the initial activity,- \(\lambda\) is the decay constant,- \(t\) is the time elapsed.

Step 2: Calculate the Decay Constant

We rearrange the formula to solve for \(\lambda\):\[\lambda = -\frac{1}{t}\ln\left(\frac{A}{A_0}\right)\]Given that \(A_0 = 7.56 \times 10^{11}\) Bq and \(A = 9.45 \times 10^{10}\) Bq with \(t = 92.4\) min:\[\lambda = -\frac{1}{92.4} \ln\left(\frac{9.45 \times 10^{10}}{7.56 \times 10^{11}}\right)\]Calculating this gives \(\lambda \approx 0.0270\) min\(^{-1}\).

Step 3: Determine the Half-Life

Use the relationship between the decay constant \(\lambda\) and the half-life \(T_{1/2}\):\[T_{1/2} = \frac{\ln(2)}{\lambda}\]Substitute \(\lambda = 0.0270\) min\(^{-1}\):\[T_{1/2} = \frac{\ln(2)}{0.0270} \approx 25.7 \text{ minutes}\]

Step 4: Calculate the Initial Number of Radioactive Nuclei

The initial activity \(A_0\) is related to the initial number of nuclei \(N_0\) by the formula:\[A_0 = \lambda N_0\]Rearranging this gives:\[N_0 = \frac{A_0}{\lambda}\]Substitute \(A_0 = 7.56 \times 10^{11}\) Bq and \(\lambda = 0.0270\) min\(^{-1}\):\[N_0 = \frac{7.56 \times 10^{11}}{0.0270} \approx 2.8 \times 10^{13}\] nuclei.

Key Concepts

Decay Constant

Understanding the decay constant is key to unraveling the nature of radioactive decay. The decay constant, denoted as \(\lambda\), is a measure of the probability of decay of a radioactive nuclide per unit time. It tells us how quickly a substance undergoes radioactive decay. A higher \(\lambda\) indicates a faster decay rate.

In our exercise, the activity of a radioactive sample decreases over time. By applying the decay formula \( A = A_0 e^{-\lambda t} \), we are able to understand how the decay process is influenced by the decay constant. Using the values provided in the problem and rearranging the formula, we calculate \(\lambda\).

Here's a quick

• Given the initial activity \(A_0\) and activity at time \(t\), we solve for \(\lambda\) using the relation:\[\lambda = -\frac{1}{t}\ln\left(\frac{A}{A_0}\right)\]
• A precise calculation of \(\lambda\) is crucial as it directly impacts other important measurements, like the half-life.

Half-Life Calculation

The half-life of a radioactive nuclide is the time it takes for half of a given sample to decay. It's an essential concept in understanding radioactive decay because it provides a practical measurement of the decay process itself.

The relationship between the decay constant and the half-life \(T_{1/2}\) is given by the formula:

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

This equation shows us that the half-life is inversely proportional to the decay constant. A smaller \(\lambda\) means a longer half-life, as the substance decays more slowly.

In the provided exercise, substituting \(\lambda = 0.0270\) min\(^{-1}\) into the equation allows us to calculate the half-life of the nuclide as approximately 25.7 minutes. This means that in this time span, the activity of the nuclide will reduce by half, providing a clear sense of the decay rate for practical applications.

Radioactive Nuclide

A radioactive nuclide is an unstable form of an element that decays, emitting radiation in the process. These decays occur naturally or can be induced in a laboratory setting.

In the specific example of \(^{199} \mathrm{Pt}\), the focus is on understanding its initial state and activity. The initial activity \(A_0\) provides a baseline for how many radioactive nuclei were initially present in the sample.

To find the initial number of radioactive nuclei \(N_0\), we use the initial activity and the decay constant:\[ A_0 = \lambda N_0 \]

Rearranging gives us:

• \[ N_0 = \frac{A_0}{\lambda} \]

For the given initial activity, \(7.56 \times 10^{11}\) Bq, and a decay constant \(\lambda = 0.0270\) min\(^{-1}\), we find that \( N_0 \) is roughly \(2.8 \times 10^{13}\) nuclei. This calculation offers insight into how many unstable atoms were present initially, setting the stage for how this nuclide will decay over time.