Question

The lattices of Na and Al are bcc and fcc respectively. Presuming them to be closed packed, their packing fractions are, respectively :

• A. $0.68$ and $0.34$
• B. $0.68$ and $0.74$
• C. $0.34$ and $0.34$
• D. $0.52$ and $0.52$

Answer 1

The correct option is B $0.68$ and $0.74$

Packing fraction is known as the fraction of volume occupied by the constituent particles in crystal structure.

GIVEN

Lattices of $Na$ is Body centred cubic $BCC$

Lattices of $Al$ is Face centred cubic $FCC$

SOLUTION

$Packingfraction$ =$\frac{N_{atom}{V}_{atom}}{V_{unitcell}}$

Packing fraction of BCC ($Na$)

Number of atoms ($N_{atom}$) = $2$

Volume of atoms ($V_{atom}$) = $\frac{4}{3}$$\pi$$r^2$

Volume of unit cell ($V_{unitcell}$) = $($$\frac{4r}{\sqrt{3}}$${)}^3$

Particle fraction ${}_{BCC}$ =$\frac{N_{atom}{V}_{atom}}{V_{unitcell}}$

= $\frac{2\times \frac{4}{3}\pi r^2}{(\frac{4r}{\sqrt{3}}{)}^3}$

= $\frac{\pi \sqrt{3}}{8}$

$Particlefraction$${}_{BCC}$$\cong$ $0.6802$

Packing fraction of FCC ($Al$)

Number of atoms ($N_{atom}$) = $4$

Volume of atoms ($V_{atom}$) = $\frac{4}{3}$$\pi$$r^{30}$

Volume of unit cell ($V_{unitcell}$) = $($$\frac{4r}{\sqrt{2}}$${)}^3$

Particle fraction ${}_{FCC}$ = $\frac{4\times \frac{4}{3}\pi r^3}{(\frac{4r}{\sqrt{2}}{)}^3}$

=$\frac{\pi \sqrt{2}}{12}$

Particle fraction ${}_{FCC}$ = $0.74$

The particle fractions for $Na$ and $Al$ are $0.68$ and $0.74$.

The correct option : B