Question

Element A has a face centred unit cell in which A is absent from half of the face centres. The packing fraction is $\frac{x\times \sqrt{2}\pi }{96}$.
The value of x is :

Answer 1

$\text{Contribution for a corner particle is}=\frac{1}{8}$
$\text{Contribution for a face centre particle is}=\frac{1}{2}$

Total no. of particles :

$Z_{eff}=(8\times \frac{1}{8})+(3\times \frac{1}{2})=\frac{5}{2}$
For a FCC unit cell :
$\text{Edge length}\left(a\right)=2\sqrt{2}\times r$

Here r is the radius of particle A
$\text{Total volume}V=a^3=(2\sqrt{2}\times r{)}^{3}V=16\sqrt{2}\times r^3$
$\text{Volume occupied}\left(V_o\right)=Z_{eff}\times \frac{4}{3}\times \pi \times r^3{V}_o=\frac{5}{2}\times \frac{4}{3}\times \pi \times r^3$
$\text{Packing Fraction (P.F)}=\frac{V_o}{V}P.F=\frac{\frac{5}{2}\times \frac{4}{3}\times \pi \times r^3}{16\sqrt{2}\times r^3}$

$P.F=\frac{10\pi }{48\sqrt{2}}P.F=\frac{10\sqrt{2}\pi }{48\times \sqrt{2}\times \sqrt{2}}=\frac{10\sqrt{2}\pi }{96}$

So value of x is 10