The activity of a radioactive sample is `1.6` curie, and its half-life is `2.5 days`. Its activity after `10 days` will be

Options

A. `0.8 curie`
B. `0.4 curie`
C. `0.1 curie`
D. `0.16curie`

Explanation

To find the activity of a radioactive sample after a certain period, we can use the formula derived from the disintegration law. Here are the steps to solve the problem:

### Step 1: Understand the Given Data
- Initial activity \( A_0 = 1.6 \) Curie
- Half-life \( t_{1/2} = 2.5 \) days
- Time elapsed \( t = 10 \) days

### Step 2: Calculate the Decay Constant (\( \lambda \))
The decay constant \( \lambda \) can be calculated using the half-life formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
Substituting the half-life:
\[ \lambda = \frac{\ln(2)}{2.5} \]

### Step 3: Calculate the Activity After 10 Days
The activity after time \( t \) can be calculated using the formula:
\[ A(t) = A_0 e^{-\lambda t} \]
Substituting the values we have:
\[ A(10) = 1.6 \cdot e^{-\lambda \cdot 10} \]

### Step 4: Substitute the Value of \( \lambda \)
Now, substituting \( \lambda \) into the equation:
\[ A(10) = 1.6 \cdot e^{-\left(\frac{\ln(2)}{2.5}\right) \cdot 10} \]
This simplifies to:
\[ A(10) = 1.6 \cdot e^{-4 \ln(2)} = 1.6 \cdot e^{\ln(2^{-4})} = 1.6 \cdot 2^{-4} \]

### Step 5: Calculate \( 2^{-4} \)
Calculating \( 2^{-4} \):
\[ 2^{-4} = \frac{1}{16} \]

### Step 6: Final Calculation
Now substituting back:
\[ A(10) = 1.6 \cdot \frac{1}{16} = \frac{1.6}{16} = 0.1 \text{ Curie} \]

### Conclusion
The activity after 10 days will be \( 0.1 \) Curie.

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