A radioactive sample’s activity drops to its one-third value in 80 minutes. Calculate the half-life of the sample.

Answer

To solve the problem, we need to determine the half-life of a radioactive sample given that its activity drops to one-third of its initial value in 80 minutes. We can use the relationship between activity, time, and half-life in radioactive decay.

### Step-by-Step Solution:

1. **Understand the Relationship**:
The activity \( A \) of a radioactive sample at time \( t \) can be expressed as:
\[ A = A_0 e^{-\lambda t} \]
where \( A_0 \) is the initial activity, \( \lambda \) is the decay constant, and \( t \) is the time elapsed.

2. **Set Up the Equation**:
According to the problem, the activity drops to one-third of its initial value after 80 minutes:
\[ A = \frac{A_0}{3} \]
Substituting this into the equation gives:
\[ \frac{A_0}{3} = A_0 e^{-\lambda \cdot 80} \]

3. **Simplify the Equation**:
We can divide both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ \frac{1}{3} = e^{-\lambda \cdot 80} \]

4. **Take the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{3}\right) = -\lambda \cdot 80 \]

5. **Calculate the Decay Constant \( \lambda \)**:
We know that:
\[ \ln\left(\frac{1}{3}\right) = -\ln(3) \approx -1.0986 \]
Thus, we have:
\[ -1.0986 = -\lambda \cdot 80 \]
Rearranging gives:
\[ \lambda = \frac{1.0986}{80} \approx 0.0137325 \text{ min}^{-1} \]

6. **Relate \( \lambda \) to Half-Life**:
The half-life \( t_{1/2} \) is related to the decay constant \( \lambda \) by the formula:
\[ t_{1/2} = \frac{\ln(2)}{\lambda} \]
where \( \ln(2) \approx 0.693 \).

7. **Calculate the Half-Life**:
Substituting the value of \( \lambda \):
\[ t_{1/2} = \frac{0.693}{0.0137325} \approx 50.46 \text{ minutes} \]

### Final Answer:
The half-life of the sample is approximately **50.46 minutes**.