Let `N_(beta)` be the number of `beta` particles emitted by `1` gram of `Na^(24)` radioactive nuclei (half life `= 15` hrs) in `7.5` hours, `N_(beta)` is close to (Avogadro number `= 6.023 xx 10^(23) // "g. mole"`) :-

Options

A. `1.75 xx 10^(22)`
B. `6.2 xx 10^(21)`
C. `7.5 xx 10^(21)`
D. `1.25 xx 10^(22)`

Answer

To solve the problem, we need to calculate the number of beta particles emitted by 1 gram of \( \text{Na}^{24} \) nuclei in 7.5 hours, given that the half-life of \( \text{Na}^{24} \) is 15 hours.

### Step-by-Step Solution:

1. **Determine the Initial Number of Nuclei**:
The initial number of nuclei, \( N_0 \), can be calculated using the formula:
\[ N_0 = \frac{\text{mass}}{\text{molar mass}} \times N_A \]
where:
- mass = 1 gram
- molar mass of \( \text{Na}^{24} \) = 24 g/mol
- \( N_A \) (Avogadro's number) = \( 6.023 \times 10^{23} \, \text{g/mol} \)

Therefore,
\[ N_0 = \frac{1 \, \text{g}}{24 \, \text{g/mol}} \times 6.023 \times 10^{23} \approx 2.51 \times 10^{22} \, \text{nuclei} \]

2. **Calculate the Decay Constant \( \lambda \)**:
The decay constant \( \lambda \) is related to the half-life \( t_{1/2} \) by the formula:
\[ \lambda = \frac{\ln 2}{t_{1/2}} \]
Given \( t_{1/2} = 15 \, \text{hours} \):
\[ \lambda = \frac{\ln 2}{15 \, \text{hours}} \approx 0.0462 \, \text{hours}^{-1} \]

3. **Calculate the Number of Nuclei Remaining After 7.5 Hours**:
The number of nuclei remaining after time \( t \) can be calculated using the formula:
\[ N(t) = N_0 e^{-\lambda t} \]
For \( t = 7.5 \, \text{hours} \):
\[ N(7.5) = N_0 e^{-\lambda \cdot 7.5} = N_0 e^{-\frac{\ln 2}{15} \cdot 7.5} = N_0 e^{-\ln 2 \cdot 0.5} = N_0 \cdot \frac{1}{\sqrt{2}} \]

4. **Calculate the Number of Beta Particles Emitted**:
The number of beta particles emitted, \( N_{\beta} \), is the difference between the initial number of nuclei and the remaining number of nuclei:
\[ N_{\beta} = N_0 - N(7.5) = N_0 - N_0 \cdot \frac{1}{\sqrt{2}} = N_0 \left(1 - \frac{1}{\sqrt{2}}\right) \]

5. **Substituting \( N_0 \)**:
Now substituting the value of \( N_0 \):
\[ N_{\beta} = 2.51 \times 10^{22} \left(1 - \frac{1}{\sqrt{2}}\right) \]
Calculating \( 1 - \frac{1}{\sqrt{2}} \approx 0.2929 \):
\[ N_{\beta} \approx 2.51 \times 10^{22} \times 0.2929 \approx 7.36 \times 10^{21} \]

### Final Answer:
Thus, the number of beta particles emitted by 1 gram of \( \text{Na}^{24} \) in 7.5 hours is approximately:
\[ N_{\beta} \approx 7.36 \times 10^{21} \]