The half-life of a radioactive substance is 3h and its activity is `1mu Ci`. Then the activity after 9h will be (in `mu Ci`)-

Options

A. `1/9`
B. `1/27`
C. `1/3`
D. `1/8`

Explanation

To solve the problem, we need to determine the activity of a radioactive substance after a certain period of time, given its half-life and initial activity. Let's break down the steps:

### Step 1: Understand the Given Information
- The half-life (\(t_{1/2}\)) of the radioactive substance is 3 hours.
- The initial activity (\(A_0\)) is 1 µCi.
- We want to find the activity after 9 hours.

### Step 2: Determine the Number of Half-Lives
To find out how many half-lives fit into 9 hours, we divide the total time by the half-life:
\[ \text{Number of half-lives} = \frac{\text{Total time}}{t_{1/2}} = \frac{9 \text{ hours}}{3 \text{ hours}} = 3 \]

### Step 3: Calculate the Remaining Activity
The activity of a radioactive substance after a certain number of half-lives can be calculated using the formula:
\[ A = A_0 \left(\frac{1}{2}\right)^n \]
where \(n\) is the number of half-lives.

Substituting the values we have:
\[ A = 1 \, \mu Ci \left(\frac{1}{2}\right)^3 \]

### Step 4: Simplify the Expression
Calculating \(\left(\frac{1}{2}\right)^3\):
\[ \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]
Now substituting this back into the activity equation:
\[ A = 1 \, \mu Ci \times \frac{1}{8} = \frac{1}{8} \, \mu Ci \]

### Step 5: Final Answer
Thus, the activity after 9 hours will be:
\[ A = \frac{1}{8} \, \mu Ci \]

### Summary
The activity of the radioactive substance after 9 hours is \(\frac{1}{8} \, \mu Ci\).

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