A solid has a BCC structure. If the distance of the nearest approach between two atoms is 1.73 $\stackrel{\mathrm{o}}{\mathrm{A}}$. What would be the edge length of the cell?

The correct option is B

200 pm

Explanation for the correct answer:-

Option (B) 200 pm

Step 1: Calculating the edge length

In a BCC structure, the atoms are present on the corners of the cubic unit cell and one on the body center.

Such that the atoms on the corners and body center are touching each other.

Let us assume that,

Edge length of cube = a

Radius of each atom = r

$\therefore$The distance of nearest approach between two atoms is

$2\mathrm{r}=1.73\mathrm{A}$

Also, we have

$$\begin{gathered} \sqrt{3}a=4r \\ \Rightarrow a=\frac{4r}{\sqrt{3}} \\ \Rightarrow a=\frac{2\times 2r}{\sqrt{3}} \\ \Rightarrow a=\frac{2\times 1.73}{\sqrt{3}}\left[\because 2\mathrm{r}=1.73\mathrm{A}\right] \\ \Rightarrow a=2\stackrel{\mathrm{o}}{\mathrm{A}} \\ \Rightarrow a=200\mathrm{pm}\left[\because 1\stackrel{\mathrm{o}}{\mathrm{A}}=100\mathrm{pm}\right] \end{gathered}$$

Therefore, the edge length is an option (B) 200 pm.