Question

A solid element exists in simple cubic crystal. If its atomic radius is $1.0\stackrel{o}{A}$ and the ratio of packing fraction to density is $0.1c{m}^3/g,$ then the atomic mass of the element is $(N_A\approx 6\times {10}^{23})$

• A. $8\pi$
• B. $16\pi$
• C. $6\pi$
• D. $4\pi$

Answer 1

The correct option is A $8\pi$

Given element exist in simple cubic crystal
$\therefore$
$Z_{eff}=1$
Atomic radius, $r=1.0\stackrel{o}{A}={10}^{-8}cm$
For sc,
$\text{Packing fraction}=\frac{Z\times \text{Volume occupied by sphere}}{\text{Volume of the cube}}$
$\text{Packing Fraction, P.F.}=\frac{Z\times \frac{4}{3}\pi r^3}{a^3}$

$\text{Density,}\rho =\frac{Z\times M}{a^3\times N_A}$
where, $a$ is edge length of unit cell.

$\text{Ratio of}\frac{P.F.}{\rho }=\frac{4\pi r^3\times N_A}{3\times M}=0.1$
$\Rightarrow \frac{4\pi \times ({10}^{-8}{)}^3\times 6\times {10}^{23}}{3\times M}=0.1$
$M=8\pi$