Q. The density for a fcc lattice is ($A =$atomic wt., $N =$Avogadro’s number, $a =$lattice parameter)
A) $\frac{4A}{Na^{3}}$
B) $\frac{2A}{Na^{3}}$
C) $\frac{A}{Na^{3}}$
D) $\frac{A}{Na^{2}}\ $
Explanation
(a) In fcc lattice, the no. of atoms per unit cell - 4
$Density = \frac{mass\ of\ unit\ cell}{volume\ of\ unit\ cell}$
$= \frac{4A/N}{a^{3}} = \frac{4A}{Na^{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.