$$^{238}_{92}U$$ is known to undergo radioactive decay to form $$^{206}_{82}Pb$$ by emitting alpha and beta particles. A rock initially contained $$68 \times 10^{−6}$$ g of $$^{238}_{92}U$$ If the number of alpha particles that it would emit during its radioactive decay of $$^{238}_{92}U$$ to $$^{206}_{82}Pb$$ in three half-lives is Z \times 10^{18}, then what is the value of Z?

The decay series $$^{238}_{92}U \rightarrow {}^{206}_{82}Pb$$ involves successive emissions of $$\alpha$$ and $$\beta^-$$ particles.
Let the number of $$\alpha$$ particles be $$n$$ and the number of $$\beta^-$$ particles be $$m$$.

Mass-number balance: $$238-4n = 206 \;\; \Rightarrow \;\; 4n = 32 \;\; \Rightarrow \;\; n = 8$$
Atomic-number balance: $$92-2n+m = 82 \;\; \Rightarrow \;\; 92-16+m = 82 \;\; \Rightarrow \;\; m = 6$$

Thus each $$^{238}_{92}U$$ nucleus emits $$8$$ $$\alpha$$ particles during its complete transformation to $$^{206}_{82}Pb$$.

Initial mass of uranium in the rock $$= 68 \times 10^{-6}\,\text{g}$$.
Moles of $$^{238}U$$ $$= \dfrac{68 \times 10^{-6}}{238} \;\text{mol}$$.

Number of nuclei initially present:
$$N_0 = \dfrac{68 \times 10^{-6}}{238}\times 6.022 \times 10^{23}$$ $$N_0 \approx 2.857 \times 10^{-7}\times 6.022 \times 10^{23}$$ $$N_0 \approx 1.72 \times 10^{17}$$

The rock is observed for three half-lives. After $$n$$ half-lives the undecayed fraction is $$\left(\dfrac12\right)^n$$, so after three half-lives the undecayed fraction is $$\left(\dfrac12\right)^3 = \dfrac18$$.

Fraction that has decayed $$= 1-\dfrac18 = \dfrac78$$.
Number of nuclei that have decayed $$= N_0 \times \dfrac78$$.

Total $$\alpha$$ particles emitted:
$$N_\alpha = 8 \times N_0 \times \dfrac78 = 7N_0$$ $$N_\alpha = 7 \times 1.72 \times 10^{17} \approx 1.204 \times 10^{18}$$

Writing $$N_\alpha = Z \times 10^{18}$$ gives $$Z \approx 1.204$$.

Therefore, the required value is 1.204.