Question

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

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

Answer 1

Answer: (D)

Given, Density of element $(d)=10.3gc{m}^{-3}$

Edge length $(a)=314pm$

Avogadro number $\left(N_A\right)=6.02\times 10^{23}mo{l}^{-1}$

Molar mass $(M)=247gmo{l}^{-1}$

$[\because Z=96$, 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

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

$Z=\frac{10.3gc{m}^{-3}\times (314{)}^3\times 10^{-30}cm\times 6.02\times 10^{23}mo{l}^{-1}}{247gmo{l}^{-1}}$

$Z=0.78\approx 1$

Thus, the structure of solid is simple cubic.