Question

The number of atoms in 100 g of an F.C.C. crystal with density $d=10gc{m}^{-3}$ and cell edge as 200 pm is equal to:

• A. $3\times {10}^{25}$
• B. $5\times {10}^{24}$
• C. $1\times {10}^{25}$
• D. $2\times {10}^{25}$

Answer 1

The correct option is B $5\times {10}^{24}$

The volume of a cube is $V=a^3$

Density $=10g{cm}^{-3}$

No of atoms in $FCC$ unit cell is $1+6\times 0.5=4$

$D=\frac{nA}{N_{0}V}$

$\Rightarrow N_0=\frac{Z.A}{D.a^3}$

$=\frac{4\times 100}{10\times {(2\times {10}^{-8})}^3}$

$=5\times {10}^{24}$ atoms.

Hence, the correct answer is option $\text{B}$.