Number of Alpha Particles Emitted in Radioactive Decay of Uranium-238

1. Nuclear Decay Equation:
   The decay of Uranium-238 to Lead-206 is represented as:
   $^{238}_{92}\text{U} \longrightarrow\ ^{206}_{82}\text{Pb} + x\,^{4}_{2}\text{He} + y\,^{0}_{-1}\beta$
   Equating mass numbers:
   $238 = 206 + 4x \implies 4x = 32 \implies x = 8$
   Thus, each decaying $^{238}_{92}\text{U}$ nucleus emits $8$ $\alpha$-particles.

2. Decayed Moles of $^{238}_{92}\text{U}$:
   The initial mass of $^{238}_{92}\text{U}$ is $68 \times 10^{-6}\text{ g}$.
   $\text{Initial moles } n_0 = \frac{68 \times 10^{-6}\text{ g}}{238\text{ g mol}^{-1}} = \frac{2}{7} \times 10^{-6}\text{ mol}$
   (using $34 \times 2 = 68$ and $34 \times 7 = 238$).

After three half-lives ($3t_{1/2}$), the remaining fraction of $^{238}_{92}\text{U}$ is:
$\left(\frac{1}{2}\right)^3 = \frac{1}{8}$
The decayed fraction is:
$1 - \frac{1}{8} = \frac{7}{8}$
Therefore, the decayed moles of $^{238}_{92}\text{U}$ are:
$n_{\text{decayed}} = \frac{7}{8} \times \left(\frac{2}{7} \times 10^{-6}\right) = \frac{1}{4} \times 10^{-6}\text{ mol} = 0.25 \times 10^{-6}\text{ mol}$

1. Number of $\alpha$-Particles Emitted:
   $N_{\alpha} = 8 \times n_{\text{decayed}} \times N_A$
   $N_{\alpha} = 8 \times (0.25 \times 10^{-6}\text{ mol}) \times (6.022 \times 10^{23}\text{ mol}^{-1}) = 2 \times 6.022 \times 10^{17} = 1.2044 \times 10^{18}$

Comparing with $Z \times 10^{18}$:
$Z \approx 1.20$