The atomic volume for a solid crystalline metal is $(19.2 \times \sqrt{2} ) {\mathrm{cm}}^3/\mathrm{mol}$. The maximum atomic radius $(\mathrm{in} \mathrm{A}°)$ for the metal atom is

$$\begin{gathered} \mathrm{d} =\frac{\mathrm{ZM}}{{\mathrm{N}}_{\mathrm{A}} {\mathrm{a}}^3} \mathrm{g}/{\mathrm{cm}}^3 \\ \frac{1}{\mathrm{d}} = \frac{{\mathrm{N}}_{\mathrm{A}} {\mathrm{a}}^3}{\mathrm{ZM}} {\mathrm{cm}}^2 / \mathrm{gm} \\ \mathrm{atomic} \mathrm{volume} \left(\frac{1}{\mathrm{d}}\right) = \frac{{\mathrm{N}}_{\mathrm{A}} {\mathrm{a}}^3}{\mathrm{Z}} {\mathrm{cm}}^2 / \mathrm{mol} \\ {\mathrm{a}}^3 =\frac{\mathrm{atomic} \mathrm{volume} \times \mathrm{Z}}{{\mathrm{N}}_{\mathrm{A}}} \\ \mathrm{Maximum} \mathrm{occupancy} \mathrm{occurs} \mathrm{in} \mathrm{FCC} \\ \mathrm{a}= 2\sqrt{2}\mathrm{r} \\ {\left[2\sqrt{2}\mathrm{r}\right]}^3 = \frac{19.2\sqrt{2}\times 4}{6.02\times {10}^{23}} \\ 16\sqrt{2}{\mathrm{r}}^3 = \frac{19.2\sqrt{2}\times 4}{6.02 \times {10}^{23}} \\ {\mathrm{r}}^3 = \frac{19.2}{4\times 6.02\times {10}^{23}} \\ {\mathrm{r}}^3 = 0.79 \times {10}^{-23} \\ {\mathrm{r}}^3 = 7.9\times {10}^{-24} {\mathrm{cm}}^3 \\ \mathrm{r} =2\times {10}^{-8 }\mathrm{cm} \\ \mathrm{r} = 2 \mathrm{A}° \end{gathered}$$