if its density is $9.6 × 10^2 Kg/m^3$, lattice constant is 4.3 $A^0$ and atomic weight is 23. - Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 5M

Year: Dec 2013

Given:-

$a = 4.3 A^0 = 4.3 x 10^{-8} cm$

Atomic Weight, A = 23

$ρ = 9.6 × 10^2 Kg/m^3 = 0.96 g/cm^3$

To Find:-

Type of Crystal Structure

Solution:-

Taking Avogadro's number, $NA = 6.023 × 10^{23}$

Using the Formula:-

$a^3 ρ = \frac{nA}{N_A}$

$ ∴ (4.3 × 10^{-8})^3 × 0.96 = \frac{n × 23}{6.023 × 10^{23}}$

$ ∴ n = 1.999 ≈ 2$

As n = 2, the structure of the atom is BCC