Calculate the specific activities of `Na^(24) & U^(235)` nuclides whose half lives are `15` hours and `7.1xx10^(8)` years respectively.

Answer

To calculate the specific activities of the nuclides \( Na^{24} \) and \( U^{235} \), we will follow these steps:

### Step 1: Understand the formula for specific activity
The specific activity \( A \) of a radioactive nuclide is given by the formula:
\[ A = \frac{\lambda N_A}{m} \]
where:
- \( \lambda \) is the decay constant,
- \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \) mol\(^{-1}\)),
- \( m \) is the molar mass of the nuclide in grams.

### Step 2: Calculate the decay constant \( \lambda \)
The decay constant \( \lambda \) is related to the half-life \( t_{1/2} \) by the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]

### Step 3: Calculate specific activity for \( Na^{24} \)
1. **Convert half-life to seconds**:
- Given half-life \( t_{1/2} = 15 \) hours:
\[ t_{1/2} = 15 \times 60 \times 60 = 54000 \text{ seconds} \]

2. **Calculate \( \lambda \)**:
\[ \lambda = \frac{\ln(2)}{54000} \approx \frac{0.693}{54000} \approx 1.28 \times 10^{-5} \text{ s}^{-1} \]

3. **Calculate specific activity \( A \)**:
- Molar mass of sodium \( m = 24 \) g/mol:
\[ A = \frac{(1.28 \times 10^{-5}) \times (6.022 \times 10^{23})}{24} \]
\[ A \approx \frac{7.71 \times 10^{18}}{24} \approx 3.21 \times 10^{17} \text{ disintegrations per gram per second} \]

### Step 4: Calculate specific activity for \( U^{235} \)
1. **Convert half-life to seconds**:
- Given half-life \( t_{1/2} = 7.1 \times 10^8 \) years:
\[ t_{1/2} = 7.1 \times 10^8 \times 365 \times 24 \times 3600 \approx 2.24 \times 10^{16} \text{ seconds} \]

2. **Calculate \( \lambda \)**:
\[ \lambda = \frac{\ln(2)}{2.24 \times 10^{16}} \approx \frac{0.693}{2.24 \times 10^{16}} \approx 3.09 \times 10^{-17} \text{ s}^{-1} \]

3. **Calculate specific activity \( A \)**:
- Molar mass of uranium \( m = 235 \) g/mol:
\[ A = \frac{(3.09 \times 10^{-17}) \times (6.022 \times 10^{23})}{235} \]
\[ A \approx \frac{1.86 \times 10^{7}}{235} \approx 7.91 \times 10^{4} \text{ disintegrations per gram per second} \]

### Final Results
- Specific activity of \( Na^{24} \): \( 3.21 \times 10^{17} \) disintegrations per gram per second
- Specific activity of \( U^{235} \): \( 7.91 \times 10^{4} \) disintegrations per gram per second