Let $$N_\beta$$ be the number of $$\beta$$ particle emitted by 1 gram of Na$$^{24}$$ radioactive nuclei having a half life of 15 h. In 7.5 h, the number $$N_\beta$$ is close to $$[N_A = 6.023 \times 10^{23}$$ mole$$^{-1}]$$

First we must find how many $$Na^{24}$$ nuclei are present in the given 1 g sample. The molar mass of the isotope is approximately $$24\ \text{g mol}^{-1}$$, so

$$ \text{Number of moles } (n)=\frac{1\ \text{g}}{24\ \text{g mol}^{-1}}=\frac{1}{24}\ \text{mol}. $$

The Avogadro constant is $$N_A=6.023\times10^{23}\ \text{mol}^{-1}$$. Therefore the initial number of radioactive nuclei is

$$ N_0 = nN_A = \frac{1}{24}\times6.023\times10^{23} = 2.5096\times10^{22}. $$

The half-life of $$Na^{24}$$ is given as $$T_{1/2}=15\ \text{h}$$. The standard decay law states

$$ N(t)=N_0\left(\frac12\right)^{t/T_{1/2}}, $$

where $$N(t)$$ is the number of nuclei left undecayed after time $$t$$. We are interested in $$t=7.5\ \text{h}$$, which is exactly one half-life divided by two:

$$ \frac{t}{T_{1/2}}=\frac{7.5\ \text{h}}{15\ \text{h}}=\frac12. $$

So

$$ N(t)=N_0\left(\frac12\right)^{1/2}=N_0\frac1{\sqrt2}=N_0\times0.7071. $$

Substituting $$N_0=2.5096\times10^{22}$$, we obtain

$$ N(t)=2.5096\times10^{22}\times0.7071 =1.775\times10^{22}. $$

The number of nuclei that have decayed in 7.5 h is therefore

$$ \Delta N = N_0 - N(t) = 2.5096\times10^{22}-1.775\times10^{22} = 0.7346\times10^{22} = 7.346\times10^{21}. $$

Each decay of $$Na^{24}$$ emits one $$\beta$$ particle, so the number of $$\beta$$ particles emitted in 7.5 h is

$$ N_\beta = \Delta N \approx 7.3\times10^{21}. $$

Among the given options, this value is closest to $$7.5\times10^{21}$$.

Hence, the correct answer is Option C.