A radioactive sample had an initial activity of 56 dpm . After 69.3 minutes, it was found to have an activity of 28 dpm . Find the number of atoms in a sample having an activity of 100 dpm.

Options

A. 693
B. 100
C. 1000
D. 10000

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the relationship between activity and the number of atoms
The activity \( A \) of a radioactive sample is given by the equation:
\[ A = \lambda N \]
where \( \lambda \) is the decay constant and \( N \) is the number of radioactive atoms in the sample.

### Step 2: Determine the decay constant \( \lambda \)
The decay constant \( \lambda \) can be calculated using the half-life \( t_{1/2} \):
\[ \lambda = \frac{0.693}{t_{1/2}} \]
From the problem, we know that the half-life \( t_{1/2} \) is 69.3 minutes.

### Step 3: Calculate \( \lambda \)
Substituting the value of \( t_{1/2} \):
\[ \lambda = \frac{0.693}{69.3 \text{ min}} = 0.01 \text{ min}^{-1} \]

### Step 4: Set up the activity equations
Initially, the activity \( A_0 \) is 56 dpm (disintegrations per minute), and after 69.3 minutes, the activity \( A \) is 28 dpm. We can express the relationship between the initial and final activities as:
\[ A = A_0 e^{-\lambda t} \]
Substituting the known values:
\[ 28 = 56 e^{-0.01 \times 69.3} \]

### Step 5: Solve for \( e^{-\lambda t} \)
Dividing both sides by 56:
\[ \frac{28}{56} = e^{-0.01 \times 69.3} \]
\[ 0.5 = e^{-0.693} \]
This confirms that the calculations are consistent since \( e^{-0.693} \approx 0.5 \).

### Step 6: Find the number of atoms for an activity of 100 dpm
Now we want to find \( N \) when \( A = 100 \) dpm:
\[ 100 = \lambda N \]
Substituting \( \lambda = 0.01 \):
\[ 100 = 0.01 N \]
\[ N = \frac{100}{0.01} = 10000 \]

### Conclusion
The number of atoms in a sample having an activity of 100 dpm is **10,000**.