A radioactive sample has a half-life of 20 years. The time at which the activity of the sample will reduce to 10% of its initial value is

Options

A. 75 y
B. 85 y
C. 66 y
D. 45 y

Explanation

To solve the problem of determining the time at which the activity of a radioactive sample reduces to 10% of its initial value, given that its half-life is 20 years, we can follow these steps:

### Step 1: Understand the concept of half-life
The half-life of a radioactive substance is the time taken for half of the radioactive nuclei in a sample to decay. In this case, the half-life is given as 20 years.

### Step 2: Write the relationship for radioactive decay
The activity of a radioactive sample can be expressed using the equation:
\[ A(t) = A_0 e^{-\lambda t} \]
where:
- \( A(t) \) is the activity at time \( t \),
- \( A_0 \) is the initial activity,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

### Step 3: Relate half-life to the decay constant
The decay constant \( \lambda \) can be related to the half-life \( t_{1/2} \) by the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Given \( t_{1/2} = 20 \) years, we can calculate \( \lambda \):
\[ \lambda = \frac{\ln(2)}{20} \]

### Step 4: Set up the equation for 10% activity
We want to find the time \( t \) when the activity reduces to 10% of its initial value:
\[ A(t) = 0.1 A_0 \]
Substituting this into the decay equation gives:
\[ 0.1 A_0 = A_0 e^{-\lambda t} \]
Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ 0.1 = e^{-\lambda t} \]

### Step 5: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ \ln(0.1) = -\lambda t \]
Since \( \ln(0.1) = -\ln(10) \):
\[ -\ln(10) = -\lambda t \]
Thus:
\[ \lambda t = \ln(10) \]

### Step 6: Substitute for \( \lambda \)
Substituting \( \lambda = \frac{\ln(2)}{20} \):
\[ \frac{\ln(2)}{20} t = \ln(10) \]

### Step 7: Solve for \( t \)
Rearranging gives:
\[ t = \frac{20 \ln(10)}{\ln(2)} \]
Now, we can calculate this value:
- \( \ln(10) \approx 2.302 \)
- \( \ln(2) \approx 0.693 \)

Substituting these values in:
\[ t \approx \frac{20 \times 2.302}{0.693} \]
Calculating this gives:
\[ t \approx \frac{46.04}{0.693} \approx 66.45 \text{ years} \]

### Conclusion
The time at which the activity of the sample will reduce to 10% of its initial value is approximately **66.45 years**.

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