Activity of radioactive element decreased to one third of original activity `R_0` in `9` years. After further `9` years, its activity will be

Options

A. `R_0`
B. `(2)/(3) R_0`
C. `R_0//9`
D. `R_0//6`

Explanation

To solve the problem, we need to calculate the activity of a radioactive element after a total of 18 years, given that its activity decreases to one third of its original activity \( R_0 \) in the first 9 years.

### Step-by-Step Solution:

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

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

3. **Simplifying the equation**:
We can divide both sides by \( R_0 \) (assuming \( R_0 \neq 0 \)):
\[ \frac{1}{3} = e^{-9\lambda} \]

4. **Taking the natural logarithm**:
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{1}{3}\right) = -9\lambda \]

5. **Finding the decay constant \( \lambda \)**:
Rearranging the equation:
\[ \lambda = -\frac{1}{9} \ln\left(\frac{1}{3}\right) \]

6. **Calculating the activity after another 9 years**:
Now, we need to find the activity after a total of 18 years. The time \( t \) is now 18 years, so we substitute \( t = 18 \) into the decay formula:
\[ R_{18} = R_0 e^{-18\lambda} \]

7. **Substituting for \( e^{-18\lambda} \)**:
We can express \( e^{-18\lambda} \) in terms of \( e^{-9\lambda} \):
\[ e^{-18\lambda} = (e^{-9\lambda})^2 \]
From step 4, we know \( e^{-9\lambda} = \frac{1}{3} \):
\[ e^{-18\lambda} = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]

8. **Final calculation of activity**:
Now substituting back into the activity formula:
\[ R_{18} = R_0 e^{-18\lambda} = R_0 \cdot \frac{1}{9} \]
Therefore, the activity after 18 years is:
\[ R_{18} = \frac{R_0}{9} \]

### Conclusion:
The activity of the radioactive element after a total of 18 years will be \( \frac{R_0}{9} \).