After 280 days, the activity of a radioactive sample is 6000 dps. The activity reduces to 3000 dps after another 140 days. The initial activity of the sample in dps is

Options

A. (a) 6000
B. (b) 9000
C. (c) 3000
D. (d) 24000

Explanation

To find the initial activity of the radioactive sample, we can use the concept of half-life and the exponential decay of radioactive substances. Here’s a step-by-step solution:

### Step 1: Understand the problem
We know that the activity of a radioactive sample decreases over time. We are given:
- After 280 days, the activity is 6000 dps.
- After another 140 days (total 420 days), the activity is 3000 dps.

### Step 2: Determine the number of half-lives
The activity reduces from 6000 dps to 3000 dps in 140 days. This means that in this period, the activity has halved. Therefore, we can conclude that:
- The time taken for one half-life (T1/2) is 140 days.

### Step 3: Calculate the number of half-lives in 280 days
Since we have 280 days:
- Number of half-lives (n) in 280 days = 280 days / 140 days = 2 half-lives.

### Step 4: Use the decay formula
The formula for radioactive decay is given by:
\[ A = A_0 \left( \frac{1}{2} \right)^n \]
where:
- \( A \) is the final activity,
- \( A_0 \) is the initial activity,
- \( n \) is the number of half-lives.

### Step 5: Substitute the known values
After 280 days (2 half-lives), the activity is 6000 dps. Therefore:
\[ 6000 = A_0 \left( \frac{1}{2} \right)^2 \]
\[ 6000 = A_0 \left( \frac{1}{4} \right) \]

### Step 6: Solve for the initial activity \( A_0 \)
Rearranging the equation gives:
\[ A_0 = 6000 \times 4 \]
\[ A_0 = 24000 \, \text{dps} \]

### Conclusion
The initial activity of the radioactive sample is **24000 dps**.