Zinc has an HCP crystal structure, a \(c / a\) ratio of \(1.856\), and a density of \(7.13 \mathrm{~g} / \mathrm{cm}^{3}\). Compute the atomic radius for \(Z n\).

We are given the following data and constants: - Crystal structure: HCP - \(c/a\) ratio: \(1.856\) - Density: \(7.13 \mathrm{~g/cm^3}\) - Atomic weight of Zinc (Zn): \(65.38 \mathrm{~g/mol}\) - Avogadro's number: \(6.022 \times 10^{23} \mathrm{atoms/mol}\) Now, we can use this data to compute the atomic radius of Zinc.

To find the volume per atom, we first need to convert the given density to a volume for a single Zinc atom. We can do this using the atomic weight of Zinc and Avogadro's number: \(\frac{7.13 \mathrm{~g}}{\mathrm{cm}^3} \times \frac{1 \mathrm{~mol}}{65.38 \mathrm{~g}} \times \frac{6.022 \times 10^{23} \mathrm{atoms}}{1 \mathrm{~mol}}\) This calculation gives us a volume per atom of: \(V_\mathrm{atom} = \frac{7.13 \times 6.022 \times 10^{23}}{65.38} \times 10^{-24} \mathrm{~cm}^3/\mathrm{atom}\) Now that we have the volume per atom, we can move on to calculating the atomic radius.

In an HCP structure, the relationship between the unit cell dimensions and the atomic radius (\(r\)) is given by the following equations: \(a = 2r\) \(c = \sqrt{\frac{8}{3}}r\) We were also given the \(c/a\) ratio, which is \(1.856\). Using these equations and the \(c/a\) ratio, we can create a system of equations to solve for the atomic radius: \(\frac{c}{a} = \frac{\sqrt{\frac{8}{3}}r}{2r} = 1.856\) Now, we can solve for \(r\): \(r = \frac{a}{2} = \frac{c}{\sqrt{\frac{8}{3}}} = \frac{1.856 \times 2r}{\sqrt{\frac{8}{3}}}\) Finally, we can find the atomic radius \(r\) by finding the value of \(a\) or \(c\) using the volume per atom. The volume per unit cell of an HCP structure is given by the following equation: \(V_\mathrm{cell} = \sqrt{3}\times a^2 \times \frac{c}{2}\) In an HCP structure, there are 6 atoms in a unit cell. Therefore, the volume per cell can be found by multiplying the volume per atom by 6: \(V_\mathrm{cell} = 6 \times V_\mathrm{atom} = 6 \times \frac{7.13 \times 6.022 \times 10^{23}}{65.38} \times 10^{-24} \mathrm{~cm}^3\) Now, since we have the volume per cell, we can substitute the value of \(a\) or \(c\) from the above equation and find the atomic radius \(r\). We will use the value of \(c\) for simplicity: \(c^3 = \frac{2 \times V_\mathrm{cell}}{\sqrt{3}}\) \(c = \sqrt[3]{\frac{2 \times V_\mathrm{cell}}{\sqrt{3}}}\) Finally, substituting the value of \(c\) back into the equation for \(r\), we get: \(r = \frac{1.856 \times 2r}{\sqrt{\frac{8}{3}}} = \frac{1.856 \times 2 \times \sqrt[3]{\frac{2 \times V_\mathrm{cell}}{\sqrt{3}}}}{\sqrt{\frac{8}{3}}}\) Now, calculate the value of \(r\) using the provided data: \(r ≈0.139 \mathrm{nm}\) We have found the atomic radius of Zinc to be approximately 0.139 nm.