Activity of a radioactive sammple decreases to (1/3)rd of its original value in 3 days. Then, in 9 days its activity will become

Options

A. `(1)/(27)` of the original value
B. `(1)/(9)` of the original value
C. `(1)/(8)` of the original value
D. `(1)/(3)` of the original value

Answer

To solve the problem, we need to determine how the activity of a radioactive sample changes over time based on its decay characteristics.

### Step-by-Step Solution:

1. **Understanding the Decay Formula**:
The activity \( R \) of a radioactive sample at time \( t \) can be expressed using the formula:
\[ R = R_0 e^{-\lambda t} \]
where:
- \( R_0 \) is the initial activity,
- \( \lambda \) is the decay constant,
- \( t \) is the time elapsed.

2. **Setting Up the Initial Condition**:
According to the problem, the activity decreases to \( \frac{1}{3} \) of its original value in 3 days. This gives us:
\[ R = \frac{1}{3} R_0 \quad \text{when} \quad t = 3 \text{ days} \]
Substituting into the decay formula:
\[ \frac{1}{3} R_0 = R_0 e^{-\lambda \cdot 3} \]

3. **Simplifying the Equation**:
Dividing both sides by \( R_0 \) (assuming \( R_0 \neq 0 \)):
\[ \frac{1}{3} = e^{-3\lambda} \]

4. **Taking the Natural Logarithm**:
To solve for \( \lambda \), we take the natural logarithm of both sides:
\[ \ln\left(\frac{1}{3}\right) = -3\lambda \]
Therefore, we can express \( \lambda \) as:
\[ \lambda = -\frac{1}{3} \ln\left(\frac{1}{3}\right) \]

5. **Finding Activity After 9 Days**:
Now we need to find the activity \( R' \) after 9 days:
\[ R' = R_0 e^{-\lambda \cdot 9} \]
We can express \( e^{-9\lambda} \) using the previously found \( e^{-3\lambda} \):
\[ e^{-9\lambda} = (e^{-3\lambda})^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]

6. **Substituting Back**:
Thus, we have:
\[ R' = R_0 \cdot \frac{1}{27} \]

7. **Final Result**:
Therefore, the activity after 9 days is:
\[ R' = \frac{1}{27} R_0 \]
This means the activity becomes \( \frac{1}{27} \) of the original value.

### Conclusion:
The correct answer is \( \frac{1}{27} \) of the original value, which corresponds to option A.