The activity of a radioative element decreases to one third of the original activity `I_(0)` in a period of nine years. After a further 1apse of nine years, its activity will be

Options

A. `I_(0)`
B. `(2)/(3)I_(0)`
C. `(I_(0))/(9)`
D. `(I_(0))/(6)`

Explanation

To solve the problem, we need to determine the activity of a radioactive element after a total of 18 years, given that its activity decreases to one third of its original value in the first 9 years.

### Step-by-Step Solution:

1. **Understanding the decay of activity**:
We know that the activity \( R \) of a radioactive element decreases over time. The relationship can be expressed as:
\[ R = R_0 e^{-\alpha t} \]
where \( R_0 \) is the initial activity, \( \alpha \) is the decay constant, and \( t \) is the time elapsed.

2. **Setting up the equation for the first 9 years**:
After 9 years, the activity decreases to one third of the original activity:
\[ R = \frac{R_0}{3} \]
Substituting this into the decay equation:
\[ \frac{R_0}{3} = R_0 e^{-9\alpha} \]

3. **Cancelling \( R_0 \)**:
Since \( R_0 \) is not zero, we can divide both sides by \( R_0 \):
\[ \frac{1}{3} = e^{-9\alpha} \]

4. **Taking the natural logarithm**:
To solve for \( \alpha \), we take the natural logarithm of both sides:
\[ \ln\left(\frac{1}{3}\right) = -9\alpha \]
This can be rewritten as:
\[ -\ln(3) = -9\alpha \]
Therefore:
\[ \alpha = \frac{\ln(3)}{9} \]

5. **Calculating the activity after a total of 18 years**:
After another 9 years (totaling 18 years), we can use the decay formula again:
\[ R = R_0 e^{-18\alpha} \]
Substituting for \( \alpha \):
\[ R = R_0 e^{-18 \cdot \frac{\ln(3)}{9}} = R_0 e^{-2\ln(3)} = R_0 \left(e^{\ln(3)}\right)^{-2} = R_0 \left(\frac{1}{3}\right)^{2} \]
Thus:
\[ R = R_0 \cdot \frac{1}{9} \]

6. **Substituting \( R_0 \) with \( I_0 \)**:
Since \( R_0 \) is the initial activity \( I_0 \):
\[ R = \frac{I_0}{9} \]

### Final Answer:
The activity after a further lapse of 9 years will be:
\[ R = \frac{I_0}{9} \]