A radioactive sample has initial activity of 28 dpm 30 minutes later its activity 14 dpm . How many atoms of nuclide were present initially?

Options

A. 2800
B. 1212
C. 528
D. 2802

Answer

To solve the problem of determining the initial number of atoms of a radioactive nuclide based on its activity at two different times, we can follow these steps:

### Step 1: Understand the relationship between activity and number of atoms
The activity \( A \) of a radioactive sample is given by the formula:
\[ A = \lambda N_0 \]
where:
- \( A \) is the activity (in disintegrations per minute, dpm),
- \( \lambda \) is the decay constant,
- \( N_0 \) is the initial number of atoms.

### Step 2: Determine the decay constant \( \lambda \)
The decay constant \( \lambda \) is related to the half-life \( T_{1/2} \) of the radioactive substance by the formula:
\[ \lambda = \frac{0.693}{T_{1/2}} \]
In this case, we need to find \( T_{1/2} \). Since we have the activity at two different times, we can use the information given to find \( \lambda \).

### Step 3: Calculate the decay constant using the activity values
We know:
- Initial activity \( A_0 = 28 \, \text{dpm} \)
- Activity after 30 minutes \( A = 14 \, \text{dpm} \)

The relationship between the activities can also be expressed as:
\[ A = A_0 e^{-\lambda t} \]
Substituting the known values:
\[ 14 = 28 e^{-\lambda \cdot 30} \]

### Step 4: Solve for \( \lambda \)
Dividing both sides by 28:
\[ \frac{14}{28} = e^{-\lambda \cdot 30} \]
\[ 0.5 = e^{-\lambda \cdot 30} \]

Taking the natural logarithm of both sides:
\[ \ln(0.5) = -\lambda \cdot 30 \]
\[ \lambda = -\frac{\ln(0.5)}{30} \]
Calculating \( \ln(0.5) \):
\[ \lambda = \frac{0.693}{30} \approx 0.0231 \, \text{min}^{-1} \]

### Step 5: Calculate the initial number of atoms \( N_0 \)
Now we can use the initial activity to find \( N_0 \):
\[ A_0 = \lambda N_0 \]
Substituting the values:
\[ 28 = 0.0231 N_0 \]
Solving for \( N_0 \):
\[ N_0 = \frac{28}{0.0231} \approx 1212 \]

### Conclusion
The initial number of atoms of the nuclide present in the sample is approximately **1212 atoms**.