Question

Each edge of a cubic unit cell is $400 \,pm$ long. If atomic mass of the element is $120$ and its density is $6 \cdot 25\, g / cm ^{3}$, the crystal lattice is : (use $N_A =6 \times 10^{23}$ )

• A. primitive
• B. body centered
• C. face centered
• D. end centered

Answer 1

Answer: (B)

$d =6.25 \,g / cm ^{3} $
$M =120 \,d$
$a =400 \,pm$
$=4 \times 10^{-8} cm$
We know, $ d=\frac{2 \times M}{a^{3} \times N_{A}}$
$\therefore 6.25 =z \times 120 $
$=\frac{2\left(4 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}{2}=2$
$\therefore $ It is body centered