A metal crystallizes with a FCC -lattice, the edge of whose unit cell is xpm . The diameter of this metal…

In the face-centred unit cell, atoms are arranged at the corners and at centered of the face diagonal.

Let radius of metal atom = $\mathrm{r}$

So, the distance $\mathrm{AC}$= $\mathrm{r}+2\mathrm{r}+\mathrm{r}=4\mathrm{r}$

The distance $\mathrm{AD}$ and $\mathrm{CD}$ = $\mathrm{x}$$$

Using the Pythagoras theorem,

$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AD}}^2+{\mathrm{CD}}^2 \\ {\mathrm{AC}}^2={\mathrm{x}}^2+{\mathrm{x}}^2=2{\mathrm{x}}^2 \\ {\left(4\mathrm{r}\right)}^2=2{\mathrm{x}}^2 \\ \sqrt{2}\mathrm{x}=4\mathrm{r} \\ \mathrm{r}=\frac{\sqrt{2}\mathrm{x}}{4}=\frac{\mathrm{x}}{2\sqrt{2}} \end{gathered}$$

So, the diameter of this metal atom would be $\frac{\mathrm{x}}{\sqrt{2}}$.