Question

3.8 Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4 g/cm3 , and an atomic weight of 192.2 g/mol.

Answer 1

We are asked to determine the radius of an iridium atom, given that Ir has an FCC crystal structure. For \(\mathrm{FCC}, n=4\) atoms/unit cell, and \(V_{C}=16 R^{3} \sqrt{2}\) (Equation 3.4). Now,

$$ \begin{array}{c} \rho=\frac{n A_{\mathrm{Ir}}}{V_{C} N_{\mathrm{A}}} \\ =\frac{n A_{\mathrm{Ir}}}{\left(16 R^{3} \sqrt{2}\right) N_{\mathrm{A}}} \end{array} $$

And solving for \(R\) from the above expression yields

$$ R=\left(\frac{n A_{\mathrm{Ir}}}{16 \rho N_{\mathrm{A}} \sqrt{2}}\right)^{1 / 3} $$

$$ \begin{array}{c} =\left[\frac{(4 \text { atoms } / \text { unit cell }(192.2 \mathrm{~g} / \mathrm{mol})}{(16)\left(22.4 \mathrm{~g} / \mathrm{cm}^{3}\right)\left(6.022 \times 10^{23} \text { atoms } / \mathrm{mol}\right)(\sqrt{2})}\right]^{1 / 3} \\ =1.36 \times 10^{-8} \mathrm{~cm}=0.136 \mathrm{~nm} \end{array} $$

Answer 2