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Step 3: Solve for the atomic radius Divide both sides of the equation by \(4\sqrt{2}\) to isolate r: \(r = \frac{492}{4\sqrt{2}}\) Calculating the value, we get: \(r \approx 173.5 \mathrm{pm}\) So, the atomic radius of the cubic closest packed lead structure is approximately 173.5 pm. #tag_title# Step 4: Calculate the density#tag_content# To calculate the density of this lead structure, we will use the formula: \(Density = \frac{mass}{volume} = \frac{mass}{a^3}\) In a cubic closest packed structure, there are four equivalent atoms per unit cell. The molar mass of lead (Pb) is 207.2 g/mol. We can determine the mass of these four atoms in the unit cell by using Avogadro's number (\(6.022 \times 10^{23} \) atoms/mol): \(mass = 4 \cdot \frac{207.2 \mathrm{g/mol}}{6.022 \times 10^{23}\, \mathrm{atoms/mol}}\) Now, we can calculate the volume of the unit cell by using the given edge length: \(volume = a^3 = (492 \mathrm{pm})^3 \) To convert the volume from cubic picometers to cubic centimeters, we can use the following conversion factor: \(1 \mathrm{cm} = 10^{12} \mathrm{pm}\) \(volume = \frac{(492 \mathrm{pm})^3}{(10^{12}\, \mathrm{pm/cm})^3} = \frac{(492 \times 10^{-12}\, \mathrm{cm})^3}{1^3\, \mathrm{cm^3}}\) Finally, we can plug the mass and volume values into the density formula: \(Density = \frac{4 \cdot \frac{207.2\, \mathrm{g/mol}}{6.022 \times 10^{23} \, \mathrm{atoms/mol}}}{\left(\frac{492 \times 10^{-12}\, \mathrm{cm}}{1\, \mathrm{cm}}\right)^3} \approx 11.34\, \mathrm{g/cm^3}\) So, the density of the cubic closest packed lead structure is approximately 11.34 g/cm³.