Question

An element has a body-centered cubic unit cell. If one of the atom from the corner is removed. Calculate the packing fraction.

• A. $\frac{10\sqrt{2}\pi }{12}$
• B. $\frac{5\sqrt{3}\pi }{64}$
• C. $\frac{15\sqrt{3}\pi }{128}$
• D. $\frac{12\sqrt{2}\pi }{10}$

Answer 1

The correct option is C $\frac{15\sqrt{3}\pi }{128}$

$\text{Contribution for a corner particle is}=\frac{1}{8}$
$\text{Contribution for a body centre particle is}=1$

Total no. of particles :

$Z_{eff}=(7\times \frac{1}{8})+1=\frac{15}{8}$
For a BCC unit cell :
$\text{Edge length}\left(a\right)=\frac{4}{\sqrt{3}}\times r$

Here r is the radius of particle A
$\text{Total volume}V=a^3={(\frac{4}{\sqrt{3}}\times r)}^{3}V=\frac{64}{3\sqrt{3}}\times r^3$
$\text{Volume occupied}\left(V_o\right)=Z_{eff}\times \frac{4}{3}\times \pi \times r^3{V}_o=\frac{15}{8}\times \frac{4}{3}\times \pi \times r^3$
$\text{Packing Fraction (P.F)}=\frac{V_o}{V}P.F=\frac{\frac{15}{8}\times \frac{4}{3}\times \pi \times r^3}{\frac{64}{3\sqrt{3}}\times r^3}$

$P.F=\frac{15\sqrt{3}\pi }{128}$