Question

A certain form of lead has a cubic closest packed structure with an edge length of 492 pm. Calculate the value of the atomic radius and the density of lead.

Short answer

The atomic radius of lead is 174 pm.

The density of lead is 11.52 g/cc.

Step-by-step solution

Given Information

The edge length of the cube is$492\mathrm{pm}\left(\frac{{10}^{-12}\mathrm{m}}{\mathrm{pm}}\right)=4.92\times {10}^{-10}\mathrm{m}$

Concept Introduction

The density of any material can be given as,

$\mathbf{Density}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{volume}}$

Step 3: Determine the atomic radius of lead

Let the atomic radius of lead be r, and the edge length be given a = 492 pm.

So,

$$\begin{gathered} 4\mathrm{r}=\mathrm{a}\sqrt{2} \\ \mathrm{r}=\frac{\mathrm{a}}{2\sqrt{2}} \\ \mathrm{r}=\frac{4.92\times {10}^{-10}\mathrm{m}}{2\sqrt{2}} \\ \mathrm{r}=1.74\times {10}^{-10}\mathrm{m}\left(\frac{1\mathrm{pm}}{{10}^{-12}\mathrm{m}}\right) \\ \mathrm{r}=174\mathrm{pm} \end{gathered}$$

Thus, the density of the lead will be,

$$\begin{gathered} \mathrm{Density}=\frac{\mathrm{mass}}{\mathrm{volume}} \\ \mathrm{Density}=\frac{\mathrm{number}\mathrm{of}\mathrm{atoms}\mathrm{in}\mathrm{a}\mathrm{unit}\mathrm{cell}\times \mathrm{molar}\mathrm{mass}}{\mathrm{Avogadro}\text{'}\mathrm{s}\mathrm{number}\times \mathrm{volume}} \\ \mathrm{Density}=\frac{4\mathrm{atoms}\times 207.200\mathrm{gm}/\mathrm{mol}}{6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mole}\times {\left(4.92\times {10}^{-10}\mathrm{m}\right)}^3} \\ \mathrm{Density}=11.52\times {10}^6\mathrm{gm}/{\mathrm{cm}}^3 \\ \mathrm{Density}=11.52\mathrm{g}/\mathrm{cc} \end{gathered}$$