A radioactive substance has half-life of 50 sec and activity of `5 xx 10^12` becquerel. The time taken for activity to drop to `1.25 xx 10^12` is `n xx 10^2` s.

Explanation

To solve the problem step by step, we will use the concepts of half-life and radioactive decay.

### Step 1: Understand the given data
- Half-life (T₁/₂) = 50 seconds
- Initial activity (A₀) = 5 × 10¹² Bq
- Final activity (A) = 1.25 × 10¹² Bq

### Step 2: Determine the decay constant (λ)
The decay constant (λ) can be calculated using the formula:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
Substituting the half-life:
\[ \lambda = \frac{\ln(2)}{50 \text{ s}} \]

### Step 3: Use the activity formula
The activity of a radioactive substance at time t can be expressed as:
\[ A = A_0 e^{-\lambda t} \]
We need to find the time (t) when the activity drops from \(5 \times 10^{12}\) Bq to \(1.25 \times 10^{12}\) Bq.

### Step 4: Set up the equation
Substituting the values into the activity formula:
\[ 1.25 \times 10^{12} = 5 \times 10^{12} e^{-\lambda t} \]

### Step 5: Simplify the equation
Divide both sides by \(5 \times 10^{12}\):
\[ \frac{1.25 \times 10^{12}}{5 \times 10^{12}} = e^{-\lambda t} \]
This simplifies to:
\[ 0.25 = e^{-\lambda t} \]

### Step 6: Take the natural logarithm
Taking the natural logarithm of both sides:
\[ \ln(0.25) = -\lambda t \]

### Step 7: Substitute λ
Substituting the value of λ:
\[ \ln(0.25) = -\left(\frac{\ln(2)}{50}\right) t \]

### Step 8: Solve for t
Now, we know that \(0.25 = \left(\frac{1}{4}\right) = \left(\frac{1}{2}\right)^2\), so:
\[ \ln(0.25) = \ln\left(\left(\frac{1}{2}\right)^2\right) = 2 \ln\left(\frac{1}{2}\right) = -2 \ln(2) \]
Thus, we can rewrite the equation as:
\[ -2 \ln(2) = -\left(\frac{\ln(2)}{50}\right) t \]
Cancelling out the negative signs and rearranging gives:
\[ t = 2 \times 50 = 100 \text{ seconds} \]

### Step 9: Express t in the required format
The problem states that the time taken for activity to drop to \(1.25 \times 10^{12}\) Bq is \(n \times 10^2\) seconds.
Thus, we have:
\[ t = 100 = 1 \times 10^2 \text{ seconds} \]
So, \(n = 1\).

### Final Answer
The value of \(n\) is \(1\).

---