After 300 days, the activity of a radioactive sample is 5000 dps (disintegrations per sec). The activity becomes 2500 dps after another 150 days. The initial activity of the sample in dps is

Options

A. 20000
B. 10000
C. 7000
D. 25000

Explanation

To find the initial activity of the radioactive sample, we can follow these steps:

### Step 1: Understand the given data
- After 300 days, the activity (A) is 5000 dps.
- After another 150 days (total 450 days), the activity becomes 2500 dps.

### Step 2: Determine the half-life
The activity drops from 5000 dps to 2500 dps in 150 days. This indicates that the half-life (T₁/₂) of the radioactive sample is 150 days because the activity is halved.

### Step 3: Calculate the number of half-lives in 300 days
Since the half-life is 150 days, we can find how many half-lives fit into 300 days:
- Number of half-lives in 300 days = 300 days / 150 days = 2 half-lives.

### Step 4: Relate the initial activity to the current activity
The relationship between the initial activity (A₀) and the activity after a certain number of half-lives can be expressed as:
\[ A = A₀ \left(\frac{1}{2}\right)^n \]
where:
- A is the activity after n half-lives,
- A₀ is the initial activity,
- n is the number of half-lives.

### Step 5: Substitute the known values
After 2 half-lives (300 days), the activity is 5000 dps:
\[ 5000 = A₀ \left(\frac{1}{2}\right)^2 \]
\[ 5000 = A₀ \cdot \frac{1}{4} \]

### Step 6: Solve for the initial activity (A₀)
To find A₀, rearrange the equation:
\[ A₀ = 5000 \cdot 4 \]
\[ A₀ = 20000 \text{ dps} \]

### Conclusion
The initial activity of the radioactive sample is **20000 dps**.

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