Question

An element A crystallizes in fcc unit cell.
Find the new packing fraction of the unit cell so obtained when a guest atom B having the largest possible size is present at the body centre in such a way that dimension of the unit cell doesn't change.
Given :
$(\sqrt{2}-1{)}^3\approx 0.07$

• A. $\frac{4.07\pi }{12\sqrt{2}}$
• B. $\frac{6.07\pi }{15\sqrt{3}}$
• C. $\frac{2.93\pi }{10\sqrt{2}}$
• D. $\frac{3.93\pi }{15\sqrt{2}}$

Answer 1

The correct option is A $\frac{4.07\pi }{12\sqrt{2}}$

For a bcc unit cell :

In reality the face centres indicated in the figure, and guest atom B are closely packed and touching each other as shown :

a is the edge length of unit cell and r is the radius of element A
Then ,
$\text{Radius of the largest guest atom}=\frac{a-2r}{2}$
We know, for a fcc unit cell :
$a=2\sqrt{2}r$
$\therefore \text{Radius of the largest guest atom}=\frac{2\sqrt{2}r-2r}{2}=(\sqrt{2}-1)r$
Total volume occupied is equal to volume occupied by element A and the guest atom
$V_{occupied}=(4\times \frac{4}{3}\pi r^3)+\left(\frac{4}{3}\pi \right(\sqrt{2}-1{)}^3{r}^3)$
$V_{occupied}=\frac{4}{3}\pi r^3(4+(\sqrt{2}-1{)}^3)$

$\text{Total volume of the unit cell}=a^3=(2\sqrt{2}r{)}^3=16\sqrt{2}{r}^3$
$P.F=\frac{V_{occupied}}{V}$

$P.F=\frac{\frac{4}{3}\pi r^3(4+(\sqrt{2}-1{)}^3)}{16\sqrt{2}{r}^3}$
$P.F=\frac{4.07\pi }{12\sqrt{2}}$