Q.Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.

Aluminium is ccp (fcc), where atoms touch along the face diagonal, giving $\sqrt2\,a=4r$, i.e. $a=2\sqrt2\,r$.

Volume of one cell $=a^3=(3.535\times10^{-8}\text{ cm})^3=4.42\times10^{-23}\text{ cm}^3$. …

Step 1 – Relate edge length to radius for ccp.

In a cubic close-packed (face-centred cubic) structure the atoms are in contact along the face diagonal, so

Step 2 – Compute the edge length.

Step 3 – Volume of one unit cell.

Convert: $a = 354\text{ pm}=3.535\times10^{-8}\text{ cm}$.

$V_{\text{cell}} = a^3 = (3.535\times10^{-8}\text{ cm})^3 = 4.42\times10^{-23}\text{ cm}^3.$ …

Use the ccp/fcc geometry: atoms touch along the face diagonal so $a=2\sqrt2\,r$. Convert $a$ to cm, cube it for the cell volume, then divide the given …

• Using the bcc relation ($\sqrt3\,a=4r$) or simple-cubic ($a=2r$) instead of the fcc face-diagonal relation $\sqrt2\,a=4r$. …