Question

Question: Calculate the ratio of the maximum radius of an interstitial atom at the center of a simple cubic unit cell to the radius of the host atom.

Short answer

The ratio of the maximum radius of an interstitial atom at the center of a simple cubic unit cell to the radius of the host atom is 0.732.

Step-by-step solution

The maximum radius of an interstitial atom at the center of a simple cubic unit cell.

In the case of scc,

Here,

$AD=a\sqrt{3}=2\left(r+R\right)$

and,

$$\begin{gathered} 2\sqrt{3}r=2\left(r+R\right) \\ \sqrt{3}r=r+R \\ r\left(\sqrt{3}-1\right)=R \end{gathered}$$

Therefore,

$$\begin{gathered} \text{R =}\frac{a\sqrt{3}}{2}\text{-}\frac{a}{2} \\ R=\frac{a\sqrt{3}-a}{2} \\ R=\frac{0.732a}{2} \end{gathered}$$

Hence,

$r_2=\frac{0.732a}{2}$

The radius of the host atom.

In scc,

${\text{r}}_1=\frac{a}{2}$

This brings the ratio as;

$\frac{r_2}{r_1}=\frac{0.732a/2}{a/2}=0.732$