To solve the problem, we will follow these steps:
### Step 1: Calculate the number of atoms present initially
Given:
- Mass of the sample, \( m = 2.2 \, \text{mg} = 2.2 \times 10^{-3} \, \text{g} \)
- Atomic mass of \( ^{11}C \) (Carbon-11), \( A = 11 \, \text{g/mol} \)
- Avogadro's number, \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \)
The number of moles of the sample can be calculated using the formula:
\[
\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.2 \times 10^{-3} \, \text{g}}{11 \, \text{g/mol}} = 2.0 \times 10^{-4} \, \text{mol}
\]
Now, we can find the number of atoms:
\[
\text{Number of atoms} = \text{Number of moles} \times N_A = 2.0 \times 10^{-4} \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol}
\]
\[
\text{Number of atoms} \approx 1.20 \times 10^{20} \, \text{atoms}
\]
### Step 2: Calculate the activity when \( 5 \, \mu g \) of the sample will be left
Given:
- Remaining mass, \( m' = 5 \, \mu g = 5 \times 10^{-6} \, \text{g} \)
Using the same atomic mass:
\[
\text{Number of moles remaining} = \frac{5 \times 10^{-6} \, \text{g}}{11 \, \text{g/mol}} \approx 4.55 \times 10^{-7} \, \text{mol}
\]
Now, find the number of atoms remaining:
\[
\text{Number of atoms remaining} = 4.55 \times 10^{-7} \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 2.74 \times 10^{17} \, \text{atoms}
\]
### Step 3: Calculate the decay constant \( \lambda \)
The decay constant \( \lambda \) can be calculated using the half-life formula:
\[
\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{1224 \, \text{s}} \approx 5.67 \times 10^{-4} \, \text{s}^{-1}
\]
### Step 4: Calculate the activity \( R \)
The activity \( R \) is given by:
\[
R = \lambda \times N
\]
Substituting the values:
\[
R = (5.67 \times 10^{-4} \, \text{s}^{-1}) \times (2.74 \times 10^{17} \, \text{atoms}) \approx 1.55 \times 10^{14} \, \text{Bq}
\]
### Final Answers:
(i) The number of atoms present initially is approximately \( 1.20 \times 10^{20} \, \text{atoms} \).
(ii) The activity when \( 5 \, \mu g \) of the sample will be left is approximately \( 1.55 \times 10^{14} \, \text{Bq} \).
---
