The activity of a radioactive sample goes down to about 6% in a time of 2 hour. The half-life of the sample in minute is about

Options

A. (a) 30
B. (b) 15
C. (c) 60
D. (d) 120

Explanation

To solve the problem, we need to determine the half-life of a radioactive sample given that its activity decreases to about 6% in 2 hours.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know that the activity of a radioactive sample decreases to 6% of its original value in 2 hours. We need to find the half-life of the sample in minutes.

2. **Convert Time to Minutes**:
Since the time given is in hours, we convert it to minutes:
\[ 2 \text{ hours} = 2 \times 60 = 120 \text{ minutes} \]

3. **Using the Decay Formula**:
The relationship between the remaining activity \( A \) and the initial activity \( A_0 \) can be expressed as:
\[ A = A_0 e^{-\lambda t} \]
Where \( \lambda \) is the decay constant and \( t \) is the time.

4. **Setting Up the Equation**:
Since the activity goes down to 6% of its original value:
\[ 0.06 A_0 = A_0 e^{-\lambda (120)} \]
Dividing both sides by \( A_0 \):
\[ 0.06 = e^{-\lambda (120)} \]

5. **Taking the Natural Logarithm**:
To solve for \( \lambda \), we take the natural logarithm of both sides:
\[ \ln(0.06) = -\lambda (120) \]
Rearranging gives:
\[ \lambda = -\frac{\ln(0.06)}{120} \]

6. **Calculating \( \lambda \)**:
We know \( \ln(0.06) \) can be calculated:
\[ \ln(0.06) \approx -2.813 \]
Therefore:
\[ \lambda = -\frac{-2.813}{120} \approx \frac{2.813}{120} \approx 0.02344 \text{ min}^{-1} \]

7. **Finding the Half-Life**:
The half-life \( T_{1/2} \) is related to the decay constant \( \lambda \) by the formula:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]
Substituting the value of \( \lambda \):
\[ T_{1/2} = \frac{\ln(2)}{0.02344} \]
Since \( \ln(2) \approx 0.693 \):
\[ T_{1/2} \approx \frac{0.693}{0.02344} \approx 29.6 \text{ minutes} \]

8. **Final Answer**:
Rounding off, the half-life of the sample is approximately:
\[ \boxed{30 \text{ minutes}} \]