The activity fo a radioactive sample decrases t `1//3` of the orginal acticity. `A_(0)` in a period of `9` year. After `9` year more, its activity will be:

Options

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

Explanation

To solve the problem step by step, we will use the concept of radioactive decay and the relationship between activity, time, and decay constant.

### Step 1: Understand the decay of activity
The problem states that the activity of a radioactive sample decreases to \( \frac{1}{3} \) of its original activity \( A_0 \) in a period of 9 years. This means that after 9 years, the activity \( A \) can be expressed as:
\[ A = \frac{A_0}{3} \]

**Hint:** Remember that the activity of a radioactive sample decreases exponentially over time.

### Step 2: Use the decay constant
The relationship between the activity and time can be expressed using the decay constant \( \lambda \). The formula relating time \( t \), decay constant \( \lambda \), and the ratio of the initial and final activity is given by:
\[ t = \frac{2.303}{\lambda} \log\left(\frac{A_0}{A}\right) \]

Substituting \( A = \frac{A_0}{3} \) into the equation:
\[ 9 = \frac{2.303}{\lambda} \log(3) \]

**Hint:** The logarithm of a fraction can be simplified using properties of logarithms.

### Step 3: Solve for the decay constant \( \lambda \)
Rearranging the equation to solve for \( \lambda \):
\[ \lambda = \frac{2.303 \log(3)}{9} \]

Calculating \( \log(3) \) (approximately 0.477):
\[ \lambda \approx \frac{2.303 \times 0.477}{9} \approx 0.122 \]

**Hint:** Make sure to use a calculator for logarithmic values to get accurate results.

### Step 4: Calculate activity after 18 years
Now, we need to find the activity after an additional 9 years, making a total of 18 years. We can use the same formula:
\[ t = \frac{2.303}{\lambda} \log\left(\frac{A_0}{A}\right) \]
Substituting \( t = 18 \) years:
\[ 18 = \frac{2.303}{0.122} \log\left(\frac{A_0}{A}\right) \]

Rearranging gives:
\[ \log\left(\frac{A_0}{A}\right) = \frac{18 \times 0.122}{2.303} \]

Calculating the right side:
\[ \log\left(\frac{A_0}{A}\right) \approx \frac{2.196}{2.303} \approx 0.955 \]

**Hint:** Remember that \( \log\left(\frac{A_0}{A}\right) \) gives you the factor by which the activity has decreased.

### Step 5: Find the final activity
Now we can find \( A \) using:
\[ \frac{A_0}{A} = 10^{0.955} \approx 9 \]
This means:
\[ A = \frac{A_0}{9} \]

**Final Answer:** After 18 years, the activity will be \( \frac{A_0}{9} \).