At time `t=0`, activity of a radioactive substance is 1600 Bq, at t=8 s activity remains 100 Bq. Find the activity at t=2 s.

Explanation

To solve the problem of finding the activity of a radioactive substance at \( t = 2 \) seconds, we will use the exponential decay formula for radioactive decay. The steps are as follows:

### Step 1: Understand the Given Information
We are given:
- Initial activity at \( t = 0 \): \( A_0 = 1600 \, \text{Bq} \)
- Activity at \( t = 8 \, \text{s} \): \( A = 100 \, \text{Bq} \)

### Step 2: Use the Exponential Decay Formula
The formula for radioactive decay 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 3: Set Up the Equation for \( t = 8 \, \text{s} \)
At \( t = 8 \, \text{s} \):
\[ 100 = 1600 e^{-8\lambda} \]

### Step 4: Simplify the Equation
Dividing both sides by 1600:
\[ \frac{100}{1600} = e^{-8\lambda} \]
\[ \frac{1}{16} = e^{-8\lambda} \]

### Step 5: Take the Natural Logarithm
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{16}\right) = -8\lambda \]
\[ \ln(1) - \ln(16) = -8\lambda \]
\[ 0 - \ln(16) = -8\lambda \]
\[ \ln(16) = 8\lambda \]

### Step 6: Find \( e^{-2\lambda} \)
We need to find the activity at \( t = 2 \, \text{s} \):
\[ A = A_0 e^{-2\lambda} \]

From the previous step, we know:
\[ e^{-8\lambda} = \frac{1}{16} \implies e^{-2\lambda} = \left(e^{-8\lambda}\right)^{\frac{1}{4}} = \left(\frac{1}{16}\right)^{\frac{1}{4}} = \frac{1}{2} \]

### Step 7: Substitute Back into the Activity Formula
Now substituting \( e^{-2\lambda} \) back into the activity formula:
\[ A = 1600 \cdot e^{-2\lambda} = 1600 \cdot \frac{1}{2} = 800 \, \text{Bq} \]

### Final Answer
The activity at \( t = 2 \, \text{s} \) is \( 800 \, \text{Bq} \).

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