Chemistry - Solid State Question with Solution | TestHub

To find the maximum radius ratio for a foreign particle in a BCC unit cell, consider the tetrahedral void.

Let $R$ be the radius of the host iron atom and $r$ be the radius of the foreign particle.

In a BCC structure, the body diagonal is $4R=a\sqrt{3}$, so $R=\frac{a\sqrt{3}}{4}$.

The largest void in a BCC structure is the tetrahedral void. The center of a tetrahedral void is at $\left(\frac{a}{4},\frac{a}{2},0\right)$ (or equivalent positions).

The distance from the center of the void to the nearest host atom (at $\left(0,0,0\right)$ or $\left(\frac{a}{2},\frac{a}{2},\frac{a}{2}\right)$) is $R+r$.

Consider the distance from the center of the void $\left(\frac{a}{4},\frac{a}{2},0\right)$ to the host atom at $\left(0,0,0\right)$.

The distance squared is $(R+r{)}^2={\left(\frac{a}{4}\right)}^2+{\left(\frac{a}{2}\right)}^2+(0{)}^2=\frac{a^2}{16}+\frac{a^2}{4}=\frac{a^2+4a^2}{16}=\frac{5a^2}{16}$.

So, $R+r=\frac{\sqrt{5}a}{4}$.

Substitute $a=\frac{4R}{\sqrt{3}}$:

$R+r=\frac{\sqrt{5}}{4}\left(\frac{4R}{\sqrt{3}}\right)=\frac{\sqrt{5}}{\sqrt{3}}R$.

Thus, $r=\left(\frac{\sqrt{5}}{\sqrt{3}}-1\right)R=\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{3}}\right)R$.

The ratio $\frac{r}{R}=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{3}}$.

We are given that this ratio is $\frac{4\sqrt{a}-a}{a}$.

So, $\frac{\sqrt{5}-\sqrt{3}}{\sqrt{3}}=\frac{4\sqrt{a}-a}{a}=\frac{4}{\sqrt{a}}-1$.

$\frac{\sqrt{5}}{\sqrt{3}}-1=\frac{4}{\sqrt{a}}-1$.

$\frac{\sqrt{5}}{\sqrt{3}}=\frac{4}{\sqrt{a}}$.

$\sqrt{a}=\frac{4\sqrt{3}}{\sqrt{5}}$.

$a=\frac{16\times 3}{5}=\frac{48}{5}$.

We need to find the value of $5a$.

$5a=5\times \frac{48}{5}=48$.