The activity of a sample reduces from `A_(0) "to" (A_(0))/(sqrt(3))` in one hour. The activity after 3 hours more will be

Options

A. `(A_(0))/(3sqrt(3))`
B. `(A_(0))/(9)`
C. `(A_(0))/(9sqrt(3))`
D. `(A_(0))/(27)`

Answer

To solve the problem, we need to determine the activity of a radioactive sample after a total of 4 hours, given that its activity reduces from \( A_0 \) to \( \frac{A_0}{\sqrt{3}} \) in the first hour.

### Step-by-Step Solution:

1. **Understanding Activity Reduction**:
The activity of a radioactive sample decreases over time according to the formula:
\[ A(t) = A_0 e^{-\lambda t} \]
where \( A(t) \) is the activity at time \( t \), \( A_0 \) is the initial activity, and \( \lambda \) is the decay constant.

2. **Finding the Decay Constant**:
From the problem, we know that after 1 hour (i.e., \( t = 1 \) hour), the activity reduces to:
\[ A(1) = \frac{A_0}{\sqrt{3}} \]
Plugging this into the activity formula gives:
\[ \frac{A_0}{\sqrt{3}} = A_0 e^{-\lambda \cdot 1} \]
Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ \frac{1}{\sqrt{3}} = e^{-\lambda} \]

3. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ -\lambda = \ln\left(\frac{1}{\sqrt{3}}\right) = -\frac{1}{2} \ln(3) \]
Thus, we find:
\[ \lambda = \frac{1}{2} \ln(3) \]

4. **Calculating Activity After 4 Hours**:
We want to find the activity after a total of 4 hours, which means we need to calculate \( A(4) \):
\[ A(4) = A_0 e^{-\lambda \cdot 4} \]
Substituting \( \lambda \):
\[ A(4) = A_0 e^{-4 \cdot \frac{1}{2} \ln(3)} = A_0 e^{-2 \ln(3)} = A_0 \left(e^{\ln(3)}\right)^{-2} = A_0 \left(\frac{1}{3}\right)^2 = \frac{A_0}{9} \]

5. **Activity After 4 Hours**:
Therefore, the activity after 4 hours is:
\[ A(4) = \frac{A_0}{9} \]

### Final Answer:
The activity after 4 hours will be \( \frac{A_0}{9} \).

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