A

Step 1: We need to calculate the number of particles emitted by the radionuclide with a half-life of 15 hours. Let's denote the initial amount of the radionuclide as \( N_0 \) and the number of particles emitted after time \( t \) is given by the equation: \( N(t) = N_0 \left(1 - e^{-\lambda t}\right) \), where \( \lambda = \frac{0.693}{T_{1/2}} \) is the decay constant and \( T_{1/2} \) is the half-life.
Step 2: Substitute \( T_{1/2} = 15 \) hours into the formula for \( \lambda \):
\( \lambda = \frac{0.693}{15} \approx 0.0462 \) hr^-1.
Step 3: For \( t = 1 \) hour, the equation becomes:
\( N(1) = N_0 \left(1 - e^{-0.0462 \times 1}\right) \).
Now, we assume \( N_0 = \text{initial particles from 1.0 } \mu g \) of \( Na^{24} \) (assuming 100% decay for calculation).
Step 4: Calculate:
\( N(1) = N_0 \left(1 - e^{-0.0462}\right) \approx N_0 (1 - 0.954) \approx 0.0462 N_0 \).
Step 5: Using Avogadro's number \( N_A = 6 \times 10^{23} \):
If 1.0 \( \mu g \) of sodium-24 corresponds to \( N_0 = \frac{1.0 \times 10^{-6} g}{24 g/mol} \times N_A \approx 2.5 \times 10^{17} \) atoms, then for particles emitted:
\( = 0.0462 \times 2.5 \times 10^{17} \approx 1.155 \times 10^{16} \).
Step 6: Therefore, total particles emitted in one hour: \( N(1) \sim 1.16 \times 10^{16} \), rounded gives approximately \( 10^{16} \) particles.
Therefore, the answer is approximately aligned with option A.