Question

A metal $M$ is crystallised in $F.C.C$ lattice. The number of unit cells in it having $2.4\times {10}^{24}$ atoms are:

• A. $6.023\times {10}^{23}$
• B. $\frac{6.023\times {10}^{23}}{2}$
• C. $2\times 6.023\times {10}^{23}$
• D. $4\times 6.023\times {10}^{23}$

Answer 1

The correct option is A $6.023\times {10}^{23}$

In FCC lattice $Z_M=4$.

$4$ atoms of $M$ is present in a unit cell of $M$.

Number of unit cell $=\frac{2.4}{4}\times {10}^{24}$ atom $=6\times {10}^{23}$ atom $\approx N_A$ (Avogadro's number)