Question

An element crystal has a density of $8570kg/m^3$. The packing efficiency is 68 %. The closest distance of approach between neighbouring atom is 2.86 $\stackrel{o}{A}$. What is the mass of one atom approximately?
Take value of avogadro constant $6.022\times {10}^{23}$

• A. 29 amu
• B. 39 amu
• C. 63 amu
• D. 93 amu

Answer 1

The correct option is D 93 amu

The packing efficiency = 68 %, means the given lattice is BCC. The closest distance of approach = 2r

$2r=2.86\stackrel{o}{A}$
$2r=\frac{\sqrt{3}a}{2}$ or $a=\frac{2\times 2.86}{\sqrt{3}}=3.30\stackrel{o}{A}=3.30\times {10}^{-8}cm$

Let atomic Wt. of the element = M
Take value of avogadro constant $6.022\times {10}^{23}$
$\therefore \frac{2\times M}{6.022\times {10}^{23}\times {\left(3.3\right)}^3\times {10}^{-24}}=8.57$

$M=8.57\times 3\times {\left(3.3\right)}^3\times 0.1=92.39\simeq 93amu$