Calculating Radioactive Decay Time Using Half-Life

Understanding radioactive decay and half-life is fundamental in nuclear physics. A radioactive isotope's activity decreases over time as its atoms decay. The half-life is a key parameter that tells us how quickly this process happens.

The question asks for the time it takes for a radioactive isotope's activity to reduce to 3.125% of its original value, given its half-life \( T \).

Understanding Half-Life

The half-life \( T \) of a radioactive substance is the time it takes for half of the radioactive nuclei in a sample to decay, or equivalently, for the activity of the sample to reduce to half of its original value.

Radioactive Decay Formula

The activity \( A \) of a radioactive sample at time \( t \) is related to its initial activity \( A_0 \) by the formula:

\( A = A_0 \left(\frac{1}{2}\right)^n \)

where \( n \) is the number of half-lives that have passed in time \( t \). The number of half-lives is related to the total time \( t \) and the half-life \( T \) by:

\( n = \frac{t}{T} \)

Solving the Problem

We are given that the final activity \( A \) is 3.125% of the original activity \( A_0 \).

First, convert the percentage to a fraction or decimal:

\( 3.125\% = \frac{3.125}{100} = 0.03125 \)

So, \( A = 0.03125 A_0 \).

Now, substitute this into the decay formula:

\( 0.03125 A_0 = A_0 \left(\frac{1}{2}\right)^n \)

We can cancel \( A_0 \) from both sides (assuming \( A_0 \neq 0 \)):

\( 0.03125 = \left(\frac{1}{2}\right)^n \)

We need to find the value of \( n \) that satisfies this equation. Let's calculate powers of \((1/2)\):

• \(\left(\frac{1}{2}\right)^1 = 0.5\)
• \(\left(\frac{1}{2}\right)^2 = 0.25\)
• \(\left(\frac{1}{2}\right)^3 = 0.125\)
• \(\left(\frac{1}{2}\right)^4 = 0.0625\)
• \(\left(\frac{1}{2}\right)^5 = 0.03125\)

From this, we see that \( n = 5 \).

The number of half-lives that have passed is 5.

Now, we can find the total time \( t \) using the relation \( n = \frac{t}{T} \):

\( 5 = \frac{t}{T} \)

Solving for \( t \):

\( t = 5 \times T \)

\( t = 5T \)

So, it will take \( 5T \) years for the activity to reduce to 3.125% of its original value.

Step-by-Step Calculation

1. Identify the initial activity \( A_0 \) and the final activity \( A \). \( A = 3.125\% \text{ of } A_0 \).
2. Convert the percentage to a decimal: \( 3.125\% = 0.03125 \).
3. Write the ratio of final to initial activity: \( \frac{A}{A_0} = 0.03125 \).
4. Use the decay formula \( \frac{A}{A_0} = \left(\frac{1}{2}\right)^n \).
5. Substitute the ratio: \( 0.03125 = \left(\frac{1}{2}\right)^n \).
6. Determine the value of \( n \) for which \((1/2)^n\) equals 0.03125. This requires calculating powers of 1/2 until the desired value is reached. \( (1/2)^5 = 0.03125 \), so \( n = 5 \).
7. Calculate the total time \( t \) using \( t = n \times T \). With \( n=5 \), \( t = 5T \).

As the table shows, after 5 half-lives (\( 5T \) years), the activity reduces to 3.125% of the original.

Revision Table: Radioactive Decay Concepts

Additional Information: Beyond Half-Life Calculation

Radioactive decay is a random process at the individual atomic level, but predictable for a large number of atoms. The decay follows an exponential law. The activity is proportional to the number of radioactive nuclei present. Besides activity, we can also calculate the number of nuclei remaining after a certain time using a similar formula: \( N = N_0 \left(\frac{1}{2}\right)^n \), where \( N_0 \) is the initial number of nuclei and \( N \) is the number remaining.

The decay constant \( \lambda \) is another important parameter related to half-life by the equation \( T = \frac{\ln(2)}{\lambda} \). Using the decay constant, the activity can also be expressed as \( A = A_0 e^{-\lambda t} \), where \( e \) is the base of the natural logarithm. Both formulas (\( A = A_0 (1/2)^n \) and \( A = A_0 e^{-\lambda t} \)) describe the same exponential decay process, just using different parameters.