Question

If an element having atomic number $96$ crystallises in cubic lattice with a density of $10.3\,g\, cm ^{-3}$ and the edge length of $314 \,pm$ then, the structure of solid is

• A. hcp
• B. fcc
• C. bcc
• D. simple cubic

Answer 1

Answer: (D)

Given, Density of element $(d)=10.3\, g\, cm ^{-3}$

Edge length $(a)=314\, pm$

Avogadro number $\left(N_{A}\right)=6.02 \times 10^{23}\, mol ^{-1}$

Molar mass $(M)=247 \,g \,mol ^{-1}$

$[\because Z=96$, having molar mass $247$, i.e. cesium $(Cs)$ ]

$\because d=\frac{Z \times M}{a^{3} \times N_{A}}$

where, $Z=$ number of lattice points in a unit cell

$\therefore Z=\frac{d \times a^{3} \times N_{A}}{M}$

$Z=\frac{10.3\, g\, cm ^{-3} \times(314)^{3} \times 10^{-30} \,cm \times 6.02 \times 10^{23} \,mol ^{-1}}{247\, g\, mol ^{-1}}$

$Z=0.78 \approx 1$

Thus, the structure of solid is simple cubic.