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# Calculate the efficiency of packing in case of a metal crystals for : body-centered cubic

$\begin{array}{1 1}68\%\\78\%\\38\%\\58\%\end{array}$

Can you answer this question?

Packing efficiency in body-centered cubic structures :
It is clear that atom at the centre will be in touch with the order two atoms diagonally arranged.
In $\Delta EFD$
$b^2=a^2+a^2=2a^2$
$b=\sqrt 2a$
Now in $\Delta AFD$
$c^2=a^2+b^2=a^2+2a^2=3a^2$
$c=\sqrt 3a$
The length of the body diagonal c is equal to 4r,where r is the radius of the sphere(atom),as all the three spheres along the diagonal touch each other.
Therefore,$\sqrt 3a=4r$
$a=\large\frac{4r}{\sqrt 3}$
Also we can write,
$r=\large\frac{\sqrt 3}{4}$$a In this type of structure,total number of atoms is 2 and their volume is 2\times \large\frac{4}{3}$$\pi r^3$
Volume of the cube $a^3=\big(\large\frac{4r}{\sqrt 3}\big)^3$
Packing efficiency =$\large\frac{\text{ Volume occupied by two spheres in the unit cell}}{\text{Total volume of the unit cell}}$$\times 100\% \Rightarrow \large\frac{2\times \Large\frac{4}{3}\normalsize \pi r^3\times 100}{\big(\Large\frac{4r}{\sqrt 3}\big)^3}$$\%$
$\Rightarrow \large\frac{(8/3)\pi r^3\times 100}{\Large\frac{64}{3\sqrt 3}\normalsize r^3}$$\%$
$\Rightarrow 68\%$
answered Jul 30, 2014