A radioactive substances has half-life of 50 sec and activity of `5 xx 10^(12)` Becquerel. Calculate the time taken for activity to drop to `1.25 xx 10^(12)`.

Answer

To solve the problem, we need to determine the time taken for the activity of a radioactive substance to drop from \(5 \times 10^{12}\) Becquerel to \(1.25 \times 10^{12}\) Becquerel, given that the half-life of the substance is 50 seconds.

### Step-by-Step Solution:

1. **Understanding the Half-Life**:
The half-life (\(t_{1/2}\)) of a radioactive substance is the time required for half of the radioactive nuclei in a sample to decay. In this case, \(t_{1/2} = 50\) seconds.

2. **Initial and Final Activity**:
- Initial activity (\(A_0\)) = \(5 \times 10^{12}\) Bq
- Final activity (\(A\)) = \(1.25 \times 10^{12}\) Bq

3. **Finding the Number of Half-Lives**:
To find how many half-lives it takes for the activity to drop from \(5 \times 10^{12}\) Bq to \(1.25 \times 10^{12}\) Bq, we can use the formula for activity after \(n\) half-lives:
\[ A = A_0 \left(\frac{1}{2}\right)^n \]
We need to find \(n\) such that:
\[ 1.25 \times 10^{12} = 5 \times 10^{12} \left(\frac{1}{2}\right)^n \]

4. **Solving for \(n\)**:
Dividing both sides by \(5 \times 10^{12}\):
\[ \frac{1.25}{5} = \left(\frac{1}{2}\right)^n \]
Simplifying the left side:
\[ 0.25 = \left(\frac{1}{2}\right)^n \]
Recognizing that \(0.25 = \left(\frac{1}{2}\right)^2\), we have:
\[ \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^2 \implies n = 2 \]

5. **Calculating the Total Time**:
Since each half-life is 50 seconds, the total time \(t\) taken for the activity to drop to \(1.25 \times 10^{12}\) Bq is:
\[ t = n \times t_{1/2} = 2 \times 50 \text{ seconds} = 100 \text{ seconds} \]

### Final Answer:
The time taken for the activity to drop to \(1.25 \times 10^{12}\) Becquerel is **100 seconds**.

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