Question

Question: Calculate the ratio of the maximum radius of an interstitial atom at the center of each face of a bcc 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 each face of a bcc unit cell to the radius of the host atom is 0.155.

Step-by-step solution

The maximum radius of an interstitial atom at the center of each face.

In the case of bcc, the interstitial site is

Here,

$$\begin{gathered} a=\frac{4\text{R}}{\sqrt{3}} \\ a=2R+2r \end{gathered}$$

So,

$2R+2r=\frac{4\text{R}}{\sqrt{3}}$

$$\begin{gathered} 2r=\frac{4\text{R}}{\sqrt{3}}-2\text{R} \\ r=\left(\frac{2}{\sqrt{3}}-1\right)\text{R} \end{gathered}$$

As, $\text{r =}\frac{a\sqrt{3}}{4}$

Therefore,

$$\begin{gathered} r=\left(\frac{2}{\sqrt{3}}-1\right)\text{R} \\ r=0.155\times \frac{a\sqrt{3}}{4} \end{gathered}$$

Hence,

$r_2=\frac{0.155\times a\sqrt{3}}{4}$

The radius of the host atom.

In bcc,

${\text{R =r}}_1=\frac{a\sqrt{3}}{4}$

This brings the ratio as;

$$\begin{gathered} \frac{r_2}{r_1}=0.155\times \frac{a\sqrt{3}/4}{a\sqrt{3}/4} \\ \frac{r_2}{r_1}=0.155 \end{gathered}$$